Sumanta Chowdhury - CLS Aipmt-15-16 XIII Phy Study-Package-7 Set-1 Chapter-24

Chapter 24

Ray Optics and Optical Instrument Instrumentss Solutions

SECTION - A

Objective Type Questions

1.

Digitital Dig al mov movie ie proj projec ecto tors rs wor workk on the the prin princi cipl ple e of (1) Reflection from micromirrors

(2)

Refraction from thin lenses

(3) Dispersion from thin prisms

(4)

Total internal reflection from optical fibres

Sol. Answer (1)

Digital movie projectors need parabolic mirrors to converge all incident rays to a point. 2.

Dayy and Da and nigh nightt sett settin ings gs for for rear rearvi view ew mirr mirror orss uses uses (1) Thin mirrors

(2)

Thick wedge shaped mirrors

(3) Convex mirrors

(4)

Concave mirrors

Sol. Answer (2)

Day and night settings for rear view mirrors. 3.

When a beam of light is incident incident on on a plane plane mirror mirror,, it is found that a real image image is formed. The incident incident beam must be (1)) Co (1 Conv nver ergi ging ng (2)) Di (2 Divverg rgin ing g (3) Par ara allllel el (4)) Forma (4 Formation tion of real real image image by a plane plane mirror mirror is impossible impossible

Sol. Answer (1)

Real images are images formed from actual intersection of light rays.

Position of image

Virtual object

In plane mirror formation of real images is shown above. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Tower, Plot No.-4, Sector-11, Sect or-11, Dwarka, New Delhi-75 Ph.011-47623456

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4.

Ray Optics and Optical Instruments

Solution of Assignment

An object object is place placed d symmetric symmetrically ally betwee between n two plane plane mirrors, mirrors, inclin inclined ed at an angle angle of 72°, 72°, then then the total total number number of images observed is (1) 5

(2) 4

(3)

2

(4)

Infinite In

Sol. Answer (2)

For number of images formed from plane mirror m

=

360

=

360

=5

72

θ

If it was at any other point the number of images ( n) = m but this is not so. Since it is placed symmetrically : n=m–1

or n = 5 – 1 or n = 4 images 5.

A person person 1.6 m tall tall is standi standing ng at the the centre centre betwee between n two walls walls three three metre metre high. high. What What is the minimu minimum m size of a plane mirror fixed on the wall in front of him, if he is to see the full height of the wall behind him? (1) 0.8 m

(2) 1 m

(3)

1.5 m

(4)

2.3 m

Sol. Answer (2) tan θ =

h1

=

d

A

h2

1.6 − h1



....(i)

d / 2



M 1 B

tan α

=

h2 d

=

1.4 − h2

....(ii)

d / 2

 

3m

M 2

1.6 m h1

3.2 =

3

and h2

C

2.8 =

h1

3

3 – h1 – h2 = 1 m 6.

While Whi le capt capturin uring g solar solar ene energy rgy for com commer mercia ciall purpos purposes es we we use use (1) Parabolic mirrors

(2) Plane mirrors

(3)

Convex mirrors

(4)

Concave mirrors

Sol. Answer (1)

While capturing solar energy for commercial purposes we use parabolic mirrors to converge the rays coming from infinity to a point. 7.

A convex convex mirro mirrorr is used used to form form an image image of of a real objec object. t. Then Then mark the the wrong wrong statem statement ent (1) The im ima age lie iess between the pole and focus

(2)

The im ima age is di dim min inis ishe hed d in siz ize e

(3) The image is erect

(4)

The image is real

Sol. Answer (4)

A convex mirror always forms a virtual image in the care of a real object. In care of a virtual object reflected rays may intersect really to make a real image. Tower, Plot No.-4, Sector-11, Sect or-11, Dwarka, New Delhi-75 Ph.011-47623456 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower,

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4.



An object object is place placed d symmetric symmetrically ally betwee between n two plane plane mirrors, mirrors, inclin inclined ed at an angle angle of 72°, 72°, then then the total total number number of images observed is (1) 5

(2) 4

(3)

2

(4)

Infinite In

Sol. Answer (2)

For number of images formed from plane mirror m

=

360

=

360

=5

72

θ

If it was at any other point the number of images ( n) = m but this is not so. Since it is placed symmetrically : n=m–1

or n = 5 – 1 or n = 4 images 5.

A person person 1.6 m tall tall is standi standing ng at the the centre centre betwee between n two walls walls three three metre metre high. high. What What is the minimu minimum m size of a plane mirror fixed on the wall in front of him, if he is to see the full height of the wall behind him? (1) 0.8 m

(2) 1 m

(3)

1.5 m

(4)

2.3 m

Sol. Answer (2) tan θ =

h1

=

d

A

h2

1.6 − h1



....(i)

d / 2



M 1 B

tan α

=

h2 d

=

1.4 − h2

....(ii)

d / 2

 

3m

M 2

1.6 m h1

3.2 =

3

and h2

C

2.8 =

h1

3

3 – h1 – h2 = 1 m 6.

While Whi le capt capturin uring g solar solar ene energy rgy for com commer mercia ciall purpos purposes es we we use use (1) Parabolic mirrors

(2) Plane mirrors

(3)

Convex mirrors

(4)

Concave mirrors

Sol. Answer (1)

While capturing solar energy for commercial purposes we use parabolic mirrors to converge the rays coming from infinity to a point. 7.

A convex convex mirro mirrorr is used used to form form an image image of of a real objec object. t. Then Then mark the the wrong wrong statem statement ent (1) The im ima age lie iess between the pole and focus

(2)

The im ima age is di dim min inis ishe hed d in siz ize e

(3) The image is erect

(4)

The image is real

Sol. Answer (4)

A convex mirror always forms a virtual image in the care of a real object. In care of a virtual object reflected rays may intersect really to make a real image. Tower, Plot No.-4, Sector-11, Sect or-11, Dwarka, New Delhi-75 Ph.011-47623456 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower,


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A con conca cave ve mi mirr rror or of fo foca call llen engt gth h f produces produces an image n times the size of the object. If the image is real then the distance of the object from the mirror is (1) (n – 1) f

(2)) (2

 (n  1)   f  n 

(3)) (3

 (n  1)   f  n 

(4)

(n + 1) f

Sol. Answer (3) f

(magnification) m =

f

− u

Focal real image m = –n –n =

−f −f − u

 u = − 9.

f (n

+

1)

u

A co conv nvex ex mi mirr rror or ha hass a fo foca call len lengt gth h f . A real object is placed at a distance f in front of it, from the pole. It produces an image at (1) In I nfinity

(2) f

(3)) (3

f /2

(4)

2f

Sol. Answer (3)

Mirror formula :

1

v

+

1

u

=

1

f

Here object is real so u is is negative 1

v

−

1

=

u

1

f

Also (u ) = f 1

v

−

1

f

1

=

f



v

=

f 2

10. An object object placed in front of a concav concave e mirror of focal focal length 0.15 m produces produces a virtual image, image, which is twice the size of the object. The position of the object with respect to the mirror is (1) –5 – 5.5 cm

(2) –6.5 cm

(3)

–7.5 cm

(4)

–8.5 cm

Sol. Answer (3) f

m =

f

− u

f = –0.15 m m = +2 (virtual image)

2

=

0.15

−

0.15

−

−

u

 = –.075 m or – 7.5 cm. 11. When a light ray ray from a rarer medium is refracted refracted into into a denser denser medium, medium, its (1) Speed increases, wavelength in increases

(2)

Speed decreases, wa wavelength in increases

(3) Speed increases, wavele len ngth decreases

(4)

Speed decreases, wavelength decreases

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Sol. Answer (4)

When a light ray goes from a rarer to denser medium by definition of refractive index, its speed creases. Its frequency is found to be invarant. So, if velocity of light in a medium is v : : v = f 

[by wave theory]

or v   So, if v decreases, decreases,  also decreases. 12. A narrow narrow,, paraxial beam beam of light is converging converging towards a point point I on a screen. A plane parallel parallel plate of glass glass of thickness t , and refractive index µ is introduced in the path of the beam. The convergence point is shifted by (1)) t (1–1/ (1 (1–1/) away

(2) t (1 (1 + 1/) away

(3)

(1 – 1/) nearer t (1

(4)

(1 + 1/) nearer t (1

Sol. Answer (1)

Longitudinal shift is given by t – –

t µ

. Hence, point of convergence shifts by the same amount.



4 



3 

13.. The length 13 length of of a vertical vertical pole pole at the the surface surface of of a lake of of water water   

 is 24 cm. Then to an under-water fish

just below the water surface the tip of the pole appears to be (1) 18 cm above the surface

(2)

24 cm above the surface

(3) 32 cm above the surface

(4)

36 cm above the surface

Sol. Answer (3) µ=

Apparent height Real height

Now, real height = 24 cm and  = 4 3

×

24

=

4 3

App Apparent he height

 Apparent height = 32 cm 14. A ray ray of light strik strikes es a glas glasss plate plate at an angle angle 60o. If the reflected and refracted rays are perpendicular to each other, the index of refraction of glass is (1)) (1

(2) 3/2

3

(3)

( 3 2)

(4)

1/2

Sol. Answer (1)

Angle N OT = = 30º this is found by geometry after putting angle between refleted and transmitted ray equal to 90º µ =

or

sin i sin r 3

µ=

2

sin 60 60 º =

2

×

sin 30 30º

N I

R

60º 60º 90º

O

3 0 º

T

1 N'



µ=

Reflected ray

Refracted ray

3

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15. A microsco microscope pe is focussed focussed on a coin lying at at the bottom of a beaker beaker.. The microscope microscope is now raised raised by 1 cm. To what depth should water be poured into the beaker so that the coin is again in focus? (The refractive index of water is

4 3

)

(1) 1 cm

(2) 4/ 4/3 cm

(3)

3 cm

(4)

4 cm

Sol. Answer (4)

Water should be poured such that shift in depth of image is 1 cm

 d   d − µ  = 1

 − 3  = 1  4 

d 1

= 4 cm d = 16.. Two tra 16 trans nspa paren rentt medi media a A and B are separated by a plane boundary. The speed of light in medium A is 2.0 × 108 ms–1 and in medium B is 2.5 × 10 8 ms–1. The critical angle for which a ray of light going from A to B suffers total internal reflection is (1) sin–1 1 1//2

(2) sin–1 2 2//5

(3)

sin–1 4 4//5

(4)

sin–1 3/4

Sol. Answer (3)

µ A =

3 × 10

8

2 × 10

8

= 1.5 ; µB =

R.I. going from A to B =

µ AB = 0.8 =



sin C sin r

3 × 10

8

2.5 × 10 µB

µ A

or

6 8

=

5

1.2 1.5

=

= 1.2

0.8

at critical angle r = = 90º

sin C sin90º

=

0.8

C

=

−1 4

sin

5

17. Which of the follo following wing pheno phenomeno menon n of light light forms forms a rainbo rainbow? w? (1) Reflection of light

(2)

Refraction

(3) Total internal reflection

(4)

Reflection as well as refraction

Sol. Answer (4)

Formation of a rainbow involves refraction to break white light into its constituent colours. It is also involves internal reflection in the drop. 18.. Whic 18 Which h of the followin following g is possible possible applic applicatio ation n of fibre optics optics? ? (1) Endoscopy

(2)

High speed internet traffic

(3) Radio, TV & Telephone signals

(4)

All of the above

Sol. Answer (4)

Total internal reflection is used in all of the above as they involve the use of optics fibres. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Tower, Plot No.-4, Sector-11, Sect or-11, Dwarka, New Delhi-75 Ph.011-47623456

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19. An object is placed at a distance of f /2 from a convex lens. The image will be (1) At one of the foci, virtual and double its size

(2)

At 3 f /2, real and inverted

(3) At 2f , virtual and erect

(4)

At f , real and inverted

Sol. Answer (1)

Lens formula 1

1

−

v

=

u

1

f

Object is real and placed at – 1

1

=

v

+

f

1

f 2

2

f

3 =

v

v

f f

=

3

20. The least distance between a point object and its real image formed by a convex lens of focal length F is (1) 2 F

(2) 3 F

(3)

4 F

(4)

Greater than 4 F

Sol. Answer (3)

(Distance between a point object and its real image) d  4 f 21. The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against each other to form a usual convex lens. The distance from this lens at which an object must be placed to obtain a real, inverted image with magnification '–1' is (1) 80 cm

(2) 40 cm

(3)

20 cm

(4)

160 cm

Sol. Answer (2)

It forms an equi-convex lens f = 0.4 m P

1 =

0.4

=

2 .5 D

Power of combination = 5 D f

=

f

=

1

P 1

m

5

or f = 20 cm Therefore, object must be placed at 2 f (= 40 cm) 22. Two thin lenses of focal lengths 20 cm and –20 cm are placed in contact with each other. The combination has a focal length equal to (1) Infinite

(2) 50 cm

(3)

60 cm

(4)

10 cm

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Sol. Answer (1)

Powers : P 1 =

1 0.2

and P 2 = –

1 0.2

= 5 D and –5 D Net power = 0

 Focal length =

1 0

=

23. If in a plano-convex lens, radius of curvature of convex surface is 10 cm and the focal length of lens is 30 cm, the refractive index of the material of the lens will be (1) 1.5

(2) 1.66

(3)

1.33

(4)

3

Sol. Answer (3) R = 10 cm f = 30 cm 1 f

1 1 = (µ − 1)  −   R ∞ 

1 30

= ( µ − 1)

1 10

4 µ=

3

24. A glass concave lens is placed in a liquid in which it behaves like a convergent lens. If the refractive indices of glass and liquid with respect to air are ag and al respectively, then (1) ag = 5 al

(2)

 > Il

(3)

a g

 < al

a g

(4)

 = 2al

a g

Sol. Answer (3)

The glass lens behaves as divergent in air which has less R.I. It will behave as convergent in a medium of higher R.I. 25. The diameter of aperture of a plano-convex lens is 6 cm and its maximum thickness is 3 mm. If the velocity of light in the material of the lens is 2 × 108 m/s, its focal length is (1) 10 cm

(2) 15 cm

(3)

30 cm

(4)

60 cm

Sol. Answer (3)

 of medium =

=

3 × 10

8

2 × 10

8

C v 6 cm

=

1.5

3 mm

Radius of curvature of lens found using geometry (R – 3)2 + (30)2 = R 2 R 2 + 9 – 6R + 900 = R 2

909 = 6R R = 151.5 mm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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or 15.15 cm 1 f0

1 1 = (µ − 1)  −   ∞ R 

1 f 0

=

0.5 ×

30 mm 3 mm

R–3

R

1 15.15

f 0 = 30.30 cm

This is distance from optical centre O. Distance from lens is f = f 0 – thickness at principal axis = 30.3 – 0.3

 f = 30 cm 26. Two plano-convex lenses of equal focal lengths are arranged as shown

The ratio of the combined focal lengths is (1) 1 : 2 : 1

(2) 1 : 2 : 3

(3)

1:1:1

(4)

2:1:2

Sol. Answer (3)

The power of lens remains the same no matter how the lenses are placed. So adding their power, the power and hence the combined focal length will remain same in all three cases. 27. When an object is at a distance u 1 and u 2 from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is (1) u 1 – u 2

(2)

u 1

 u 2

(3)

2

u 1

 u 2 2

(4)

u 1 + u 2

Sol. Answer (3) f

m=

f

+

u

For real image

For virtual image

f

–m =

f

f

,

− u 1

f −

f

−

+m =

f =

u2

f

−

u1



f

=

f

− u 1 ,

u1 + u 2 2

28. A concave lens of focal length f produces an image (1/ x ) of the size of the object. The distance of the object from the lens is (1) ( x – 1) f (2) ( x + 1)f (3) {( x – 1)/ x }f (4) {( x + 1)/ x }f Sol. Answer (1) m=

+

1

x

f f

=

+

For virtual image m =

u

−f −f + u

,

+

1 x

u = –f ( x – 1)




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29. A thin equiconvex glass lens of refractive index 1.5 has power of 5D. When the lens is immersed in a liquid of refractive index , it acts as a divergent lens of focal length 100 cm. The value of  of liquid is (1) 4/3

(2) 3/4

(3)

5/3

(4)

8/3

Sol. Answer (3)

g = 1.5 5 = ( µg

5×2

R

2 =

R

5

=

0.2 m = 20 cm

f l < 0

µ  2 =  g − 1    µl   R 

1

fl

−

1 1 − 1)  +   R R 

1 =

P = 5 D

µ  2 =  g − 1 100  µl  20 1

µg

−1= −

µ l

1.5

=

µ l

9 10

1 10

, µ l =

5 3

30. In case of displacement method of lenses, the product of magnification in both cases is (1) 1

(2) 2

(3)

Zero

(4)

Infinite

Sol. Answer (1) m1m2 = 1 which is a fact.

Here the lens is moved between the object and the screen. v and u interchange values between the two positions a clear image is formed on the screen.

If m1 =

v u

and m2 =

u v

m1m2 = 1

31. In the displacement method, a convex lens is placed in between an object and a screen. If the magnifications in the two positions are m1 and m2 and the displacement of the lens between the two positions is x , then the focal length of the lens is (1)

x m1

 m2

(2)

x m1

– m2

x

(3)

m1  m2 2

x

(4)

m1 – m2 2


10



Sol. Answer (2)

By principle of reversibility where must be symmetry in the two position a d

u

u

v v

d = u + v

Hence,

u =

d

−

2

a

a = v – u

and v =

Putting in lens equation



f

d

2

=

−

a

2

d

−

a

d

+

a

m2 =

2 +

2

d

+

a

m1 =

=

1

f

d

+

a

d

−

a

d

−

a

d

+

a

|m1 – m2| =

2

|m1 – m2| =

4d

4 da

d

a f

2

+

a

2

. ( a = x )

x

f =

m1

– m2

32. The focal length of a planoconvex glass lens is 20 cm ( g =1.5). The plane face of it is silvered. An illuminating object is placed at a distance of 60 cm from the lens on its axis along the convex side. Then the distance (in cm) of the image is (1) 20

(2) 30

(3)

40

(4)

12

Sol. Answer (4)

Net power of combination P = 2 P L + P m

where P L is power of lens P m is power of mirrors P m = 0 P L

1 =

0.2

 P = 10 D Focal length of combination ( f ) = 1

1 −

v

1

60

10

= 0.1 m or 10 cm

1

− =

10

1 =

v

60

1

1 −

10

v = 12 cm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



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33. Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that a sharp image is formed 20 cm from the mirror. When the air between the glass pieces is replaced by water (mw = 4/3), then the image formed from the mirror is at a distance of

(1) 8 cm

(2) 10 cm

(3)

6 cm

(4)

12 cm

Sol. Answer (4)

Initially it is simply a concave mirror with f = –20 cm, R = 40 cm

 Power of mirror (P m) = =

Focal length of lens =

1 8 1 0.2

=

5D

R

2(µ − 1) 40

=

2

4 ×  − 1 3 

= 60 cm Power of lens = =

1 0.5 100

D

50

 Equivalent power = 2P L + P m P =

50 6

Net focal length = =

1 P

6 50

× 100

= 12 cm 34. Yellow light is refracted through a prism producing minimum deviation. If i 1 and i 2 denote the angle of incidence and emergence for the prism, then (1) i 1 = i 2

(2) i 1 > i 2

(3)

i 1 < i 2

(4)

i 1 + i 2 = 90

Sol. Answer (1)

At angle of minimum deviation angle of emergence of prism is same as angle of incidence. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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35. At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may graze the second surface of the prism (  = 2)? (1) 30°

(2) 90°

(3)

45°

(4)

60°

Sol. Answer (2)

Let C by critical angle  = 2 =

sin i sin r

60º

i e = 90º

60º = r + C from geometry i 1

2

µ

C

sin C

1 =

=

r

=

sin

[At emergent interface]

sin i e

−1 1

2

i e

C

60º =

60º

30º

ie = 90º

r = 30º sin i

=

sin r

2

sin i = 1 i

π =

2

36. A prism of refractive index 2 has a refracting angle of 60 o. At what angle must a ray be incident on it so that it suffers a minimum deviation? (1) 30°

(2) 45°

(3)

60°

(4)

75°

Sol. Answer (2) 2, A = 60º

µ=

At for minimum deviation : i = i e and r 1 = r 2

i

Also, r 1 + r 2 = A or

r 1

r 2

i e

2r = 60º r = 30º sin i

µ =

2

sin r

=

sin i

2 sin i 1

=

2

[Putting r = 30º]

i = 45º




13

37. A glass prism of refractive index 1.5 is immersed in water of refractive index 4/3. A light ray incident normally on face AB is totally reflected at face AC if

B



A

C

(1) sin  > 8/9

(2) sin  < 2/3

(3)

sin  =

3 2

(4)

2/3 < sin   8/9

Sol. Answer (1)

g (glass) = 1.5 l (water) = 4/3 Rays enter and pass undeviated at first interface For total interface reflection at 2nd interface i c

µ  > sin−1  e   µg 

 i

c

i c

4 2 > sin−1  ×  3 3

i c

8 > sin−1   9 8

sin i c

>

9

Since i c =  from geometry sin θ >

8 9

38. A person can see clearly only up to a distance of 25 cm. He wants to read a book placed at a distance of 50 cm. What kind of lens does he require for this purpose and what must be its power? (1) Concave, – 1.0 D

(2) Convex, + 1.5 D

(3)

Concave, – 2.0 D

(4)

Convex, + 2.0 D

Sol. Answer (3)

He needs to bring the image of the object closer to 25 cm Also the image should be virtual 1

v

−

1

u

1 −

25

1

=

f 1

+

50

1 =

f

f = –50 cm P =

1 −0.5

P = 2 D Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

14



39. An astronomical telescope has an objective of focal length 100 cm and an eye piece of focal length 5 cm. The final image of a star is seen 25 cm from the eyepiece. The magnifying power of the telescope is (1) 20

(2) 22

(3)

24

(4)

26

Sol. Answer (3) f 0 = 100 cm f e = 5 cm

To form image at near point m

 1 1 = −f 0  +  D f e

1 1  = −100  +   5 25  6 = −100    25  m = –24

40. When a telescope is adjusted for normal vision, the distance of the objective from the eye-piece is found to be 80 cm. The magnifying power of the telescope is 19. What are the focal lengths of the lenses? (1) 61 cm, 19 cm

(2) 40 cm, 40 cm

(3)

76 cm, 4 cm

(4)

50 cm, 30 cm

Sol. Answer (3)

Distance between objective and eyepiece when telescope is adjusted for normal vision is given by 19 =

 f 0  = M   f 

f0 fe

....(i)

e

80 = f 0 + f e. (L = f 0 + f e)

....(ii)

Solving (i) & (ii) f 0 = 76 cm, f 0 = 4 cm

41. The focal lengths of the objective and eye lens of a telescope are respectively 200 cm and 5 cm. The maximum magnifying power of the telescope will be (1) – 40

(2) – 48

(3)

– 60

(4)

– 100

Sol. Answer (2)

Maximum magnification is at near point Magnification at near point m

=−

m

=−

f0 fe

 f  1 + D 

200 5

e

1 + 5   25 

200 = −

5

6 ×

5

m = –48 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



15

42. A convex lens forms a real image of a point object at a distance of 50 cm from the convex lens. A concave lens is placed 10 cm behind the convex lens on the image side. On placing a plane mirror on the image side and facing the concave lens, it is observed that the final image now coincides with the object itself. The focal length of the concave lens is (1) 50 cm

(2) 20 cm

(3)

40 cm

(4)

25 cm

Sol. Answer (3)

Final image coincides with the object when rays fall normal to the mirror or parallel to principal axis. For this virtual object must be at the focus of concave lens.

50 cm

Distance of virtual object from concave lens = 50 – 10 = 40 Focal length of concave lens = 40 cm. 43. A convex lens of focal length 100 cm and a concave lens of focal length 10 cm are placed coaxially at a separation of 90 cm. If a parallel beam of light is incident on convex lens, then after passing through the two lenses the beam (1) Converges

(2) Diverges

(3)

Remains parallel

(4)

Disappears

Sol. Answer (3) f = +100 cm

f = –10 cm

90 cm 100 cm

Virtual object for concave lens is at its focus. SECTION - B

Objective Type Questions

1.

If radii of curvature of both convex surfaces is 20 cm, then focal length of the lens for an object placed in air in the given arrangement is



1

=1

 =1.5 2

(1) 10 cm

(2) 20 cm

(3)

40 cm

(4)

80 cm

Sol. Answer (3) µ3 f

=

µ2 − µ1 R1

+

µ3 − µ 2 R 2

1 = 1, 2 = 1.5, 1 =

4 3

R 1 = +20 cm, R 2 = – 20 cm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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2.



A driving mirror consists of a cylindrical mirror of radius of curvature 10 cm and the length over the curved surface is 10 cm. If the eye of the driver be assumed to be at a great distance from the mirror, then field of view in radian is (1) 2.0 (2) 4.0 (3) 3.0 (4) 5.0

Sol. Answer (1) f =

 = 3.

R

2

=

5 cm 10 cm

10 =

5

 f

2 rad

Which of the following statements is correct? (1) During hot summer days, the trees and other tall objects seem to be quivering because the density of air changes in an irregular way (2) When the moon is near the horizon it appears bigger. This is due to optical illusion (3) If the critical angle for the medium of a prism is C and the angle of prism is A, there will be no emergent ray when A > 2C

(4) All of these Sol. Answer (4) All the statements are true. 4.

An isosceles prism of angle A = 30° has one of its surfaces silvered. Light rays falling at an angle of incidence 60° on the other surface retrace their path after reflection from the silvered surface. The refractive index of prism material is 30° 60°

(1) 1.414 Sol. Answer (3)

(2) 1.5

(3)

1.732

(4)

1.866

For light to retrace its path it must reflect normally in the mirror. 30°

When it does so, by geometry r = 30º µ =

5.

sin i sin r

sin60º =

sin30º

=

3

=

60°

1.732

A short linear object of length l lies along the axis of a concave mirror at a distance u from it. If v is the distance of image from the mirror then size of the image is (1) l 

v

(2)

u

l 

u v

(3)

 v  l     u 

2

(4)

 u  l     v 

2

Sol. Answer (3)

Axial magnification of a short object is given by m

=

l 1 l

v =    u 

 v  l1 =    u 

2

2

l



6.


17

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is

4 3

and the fish is 12 cm, below the surface. The radius of the circle is r

h

C C

16

(1)

cm

(2)

7

26

cm

7

36

(3)

cm

7

(4)

46

cm

7

Sol. Answer (3) r

h = 12

 = i C sin i c =

sin i c

h

1 µw

 

3 =

4

r = h tan i C r

=

3

h

16

−

9

36 r

7.

=

7

In optical fibre, refractive index of inner part is 1.68 and refractive index of outer part is 1.44. The numerical aperture of the fibre is (1) 0.5653

(2) 0.6653

(3)

0.7653

(4)

0.8653

Sol. Answer (4)

0sin  =

2

2

µ1 − µ2

(Numerical aperture)

1 = 1.68, 2 = 1.44 8.

Compare the dispersive powers of two prisms if one of them deviates the blue and red rays through 10° and 6° respectively and the second prism through 8° and 4.5°

(1) 0.69 Sol. Answer (3) Dispersive power = δ y =

(2) 0.79

(3)

0.89

(4)

0.99

δv − δ r δ y

δv + δ r 2


18

Ray Optics and Optical Instruments 10 w 1

=

=

w 1

6

−

1 =

8 8

w 2

−


2

4.5

3.5 =

6.25

6.25

0.89

≈

w 2

9.

A thin prism of angle 6° made of glass of refractive index 1.5 is combined with another prism made of glass of refractive index 1.75 to produce dispersion without deviation. Then the angle of the second prism is (1) 7°

(2) 4°

(3)

9°

(4)

5°

Sol. Answer (2)

For dispersion without deviation : A1(1 – 1) + A2(2 – 1) = 0 6(0.5) + A2(0.75) = 0 A2 = 4º 10. In a medium of refractive index 1.6 and having a convex surface has a point object in it at a distance of 12 cm from the pole. The radius of curvature is 6 cm. Locate the image as seen from air (1) A real image at 30 cm

(2)

A virtual image at 30 cm

(3) A real image at 4.28 cm

(4)

A virtual image at 4.28 cm

Sol. Answer (2) µ2

v

−

µ1

u

=

µ2 − µ1

R

1 = 1.6, 2 = 1, u = 12 cm, R = – 6 cm 1

1.6 −

v

P

12 cm

O

1 1.6 −

=

12

6

−

−

v = – 30 cm. (Virtual image)

11. A point object is situated at a distance of 36 cm from the centre of the sphere of radius 12 cm and refractive index 1.5. Locate the position of the image due to refraction through sphere. (1) 24 cm from the surface

(2)

36 cm from the centre

(3) 24 cm from the centre

(4)

Both (1) & (2)

Sol. Answer (4)

1 = 1, 2 = 1.5 12 cm

u = –24 cm, O

R = +12 cm µ2

v

−

1.5

u

=

µ2 − µ1

R

1 −

v

µ1

−24

36 cm

=

1.5 − 1 +12



v 


19

 becomes parallel to the principal axis.

On second face, 1 = 1.5 , 2 = 1, u  , R = – 12 cm µ2

v

−

1

µ1

µ2 − µ1

=

u

1.5

R

1

−

−

=

v

1.5

12

∞

−

v = 24 cm

Final image  at 24 cm from the surface and from centre of sphere. 12. A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm. An object is situated in the denser medium at a distance of 9 cm from the pole. Locate the image due to refraction in air. (1) A real image at 8 cm

(2)

A virtual image at 8 cm

(3) A real image at 4.8 cm

(4)

A virtual image at 4.8 cm

Sol. Answer (4) µ2

v

−

1

µ1

u

1.5 +

v

1

9

=

6



1



2

−12

1

=

R

0.5

P

1

−

v

µ2 − µ1

=

−

O

24

v = –4.8

13. A light ray is travelling from air to glass. The reflected and refracted rays are perpendicular to each other. If the angle of incidence in air is i the refractive index of glass is (1) sin i

(2) cos i

(3)

tan i

(4)

cot i

Sol. Answer (3)

180º = i + 90 + r i i

r = 90 – i

µ =

r

sin i sin r

  = tan i 14. A ray is incident on boundary separating glass and water. Refractive index for glass is for water is

(1)

sin

4 3

3  1  

 4 

3 2

and refractive index

critical angle for glass-air boundary is

(2)

sin

2  1  

 3 

(3)

sin

8  1  

 9 

(4)

sin

1  1  

 9 


20



Sol. Answer (3) sin i C =

µ r µ d

Where r = R.I. of rarer medium d = R.I. of denser medium



sin i C

i C

4 =

3

2 ×

3

8 = sin−1   9

15. A plot between the angle of deviation and angle of incidence is shown in figure. From the graph one can say that the prism angle is



50º

(Angle of deviation) i 38º 58º (Angle of incidence)

(1) 47º Sol. Answer (2)

(2) 46º

(3)

45º

(4)

60º

 38º and 58º can be i or e in the equation as  will be same for both values of incidence ray. 38º + 58º = A + 50º  A = 46º i+e=A+

16. In the diagram, a ray is passing through a broken prism, find angular deviation for the ray

60º

45º

60º

(1) 105º Sol. Answer (4)

(2) 30º

30º

(3)

60º

(4)

15º

From triangle interior angles A = 180º – 60º – 30º

= 90º i+e=A+



60º

45º

45º + 60º = 90 + 

  = 15º

60º

30º




21

17. A glass slab ( = 1.5) of thickness 2 cm is placed on a spot. The shift of spot if it is viewed from top (1)

2

cm

3

(2)

4

cm

(3)

3

1

cm

3

(4)

5

cm

3

Sol. Answer (1)

Shift =

=

=

 1  − µ 

t 1

 − 1  1.5 

2 1

2

cm

3

18. Ray diagram for two lenses kept at some distance given in the diagram, which of the following option is correct (f 1, f 2 = focal length, d = distance between lenses)

(1) f 1+f 2 > d

(2)

f 1+f 2 < d

(3) f 1+f 2 = d

(4)

Combination behaves like converging lens

Sol. Answer (3) f 1

f 2

O1

P

O2

d I

II

For parallel incidence image is at f 1 distance from O1. For final emergence to be parallel object for II is at distance f 2 from O2. The distance between I and II would be f 1 + f 2. 19. A ball is projected from top of the table with initial speed u at an angle of inclination q, motion of image of ball w.r.t ball u



(1) Must be projectile

(2)

Must be straight line and vertical

(3) Must be straight line and horizontal

(4)

May be straight line, depends upon value of 


22



Sol. Answer (3)  

V I0

 

=

V I

−

V 0



V I 0

u sin

u sin

 

= −u cos θ

u

iˆ + u sin j

− (u cos θ

i + u sin θ j ) O



u cos

u cos

V I 0 = −2u cos θi ˆ  

I

Straight line and horizontal 20. In displacement method, convex lens forms a real image of an object for its two different positions. If heights of the images in two cases be 24 cm and 6 cm, then the height of the object is (1) 3 cm

(2) 36 cm

(3)

6 cm

(4)

12 cm

Sol. Answer (4)

In displacement method m1m2 = 1 h1 h

×

h2 h

=

1

h

=

h1h2

h

=

24 × 6

h = 12 cm

21. Two parallel rays of red and violet colour passed through a glass slab, which of the following is correct? Red

V i o l e t

1 2

t

3 4 5 6

(1) 3 and 4 are parallel

(2) 4 and 5 are parallel

(3)

Sol. Answer (4)

1 2

the emergent ray is parallel to incident ray. t

3 4

because 1 || 2



2 || 5

2 and 5 are parallel

Red

V i o l e t

When refraction occurs through parallel glass slab, 6 || 2, 5 || 1

6 and 3 are parallel (4)

5 6




23

22. A plane glass is kept over a coloured word ‘VIBGYOR’, where colour of letters as same as the colours in white light start by letter, the letter which appears least raised is (1) R

(2) Y

(3)

O

(4)

V

Sol. Answer (1) µ=

Real depth Apparent depth

When an image is least raised its apparent depth is highest.

 R.I. is lowest which happens to be for red light. 23. The near point of a person is 75 cm. In order that he may be able to read book at a distance 30 cm. The power of spectacles lenses should be (1) –2 D

(2) +3.75 D

(3)

+2 D

(4)

+3 D

Sol. Answer (3) 1



75

1 −

−

1



f

30

1 =

−

f

1 =

50

P = +2D

24. If a lens is moved towards the object from a distance of 40 cm to 30 cm, the magnification of the image remains the same (numarically). The focal length of the lens is (1) 20 cm

(2) 15 cm

(3)

35 cm

(4)

18 cm

Sol. Answer (3) f f

+

u1

=

−f

f

+ u2

u 1 =–40 cm, u 2 =–30 cm f  0

–(f – 40) = f – 30 –f + 40 = f – 30 f = +35 cm

25. A convex lens of power +2.5 D is in contact with a concave lens of focal length 25 cm. The power of combination is (1) –1.5 D

(2) 0 D

(3)

+1.5 D

(4)

+6.5 D

Sol. Answer (1)

Power of concave lens = −

1 0.25

= –4 D

Adding combination = –4 D + 2.5 D = –1.5 D 26. For a telescope in normal adjustment, the length of telescope is found to be 27 cm. If the magnifying power of telescope, at normal adjustment is 8, the focal lengths of objective and eye piece are respectively (1) 24 cm, 3 cm

(2) 27 cm, 8 cm

(3)

12 cm, 6 cm

(4)

27 cm, 9 cm


24



Sol. Answer (1)

For normal adjustment m =

f 0 f e

and L = f 0 + f e

f 0

8=

f e

f 0 = 8f e

From 27 = f 0 + f e 27 = 8f e + f e f e = 3 cm f 0 = 24 cm

27. A converging lens of focal length 30 cm is placed in contact with another converging lens of unknown focal length, then possible value for focal length of combination is (1) 15 cm

(2) 60 cm

(3)

36 cm

(4)

–12 cm

Sol. Answer (1)

Total power of combination will be more than power of given lens and focal length will be less. 28. In the diagram the ray passing through prism is parallel to the base. Refractive index of material of prism is

60º

45º

3

(1)

(2)

3

45º

(3)

2

(4)

2

3 2

Sol. Answer (4)

By geometry r = 45º sin60º µ=

sin45º

60º

3 =

2

[Alternate angles]

×

r

2

45º

45º

45º

3 µ =

2

29. In displacement method we use a lens of focal length f and distance between object and screen is 60 cm. Possible value for focal length is (1) –15 cm

(2) 30 cm

(3)

12 cm

(4)

20 cm

Sol. Answer (3) f 

D 4

60

f



f

 15 cm

4




25



4 



3 

30. A red colour in air has wavelength 760 nm when light passes through water of refractive index  n 

,

wavelength becomes 570 nm. (wavelength of yellow colour in air is 570 nm). Then colour of red light in water is (1) Red

(2) Green

(3)

Yellow

(4)

Blue

Sol. Answer (1)

Colour of a wave depends more on frequency than wavelength as it depicts the amount of energy is carries. Since frequency and energy doesnot change it will simply remain red. 31.

is the unit vector along incident ray, which of the following must be true? n1 ˆ

(1)

n1  n2 ˆ

ˆ

= 0

(2)

n1  n3 ˆ

ˆ

n2 ˆ

= 0

along normal and (3)

n3 ˆ

is the unit vector along reflected ray, then

( n1 × n2 ) ˆ

ˆ

 n3 ˆ

= 0

(4)

( n1 × n2 ) × ˆ

ˆ

n3 ˆ

=0

Sol. Answer (3)

The reflected ray, refracted ray a incident ray and normal all lie on the same plane. Hence (3) is true. 32. A double convex lens has two surfaces of equal radii 15 cm and refractive index  = 1.5, its focal length is equal to (1) –15 cm

(2) –30 cm

(3)

+15 cm

(4)

+30 cm

Sol. Answer (3) 1 f 1 f

1 1 = (µ − 1)  −   R −R  2 = 0.5    15 

 f = 15 cm 33. The distance between a real object and its real image is 56 cm formed by converging lens, focal length of lens is (1) f  14

(2) f > 14

(3)

f = 14

(4)

f = 28

Sol. Answer (1) f



f



D 4

56 4

 14 cm 34. In displacement method, there are two position of a lens for which we get clear image. First position of the lens is at 40 cm from object and second is at 80 cm, the focal length of lens is f

(1) 40 cm

(2)

40 3

cm

(3)

80 cm

(4)

80

cm

3


26



Sol. Answer (4)

Displacement method is based on principle of reversibility. The image distance for 1st position of lens will be same as object position with IInd position of lens. u = – 40 cm V = + 80 cm 1 f f

1 =

1

80 80

=

−

40

−

cm

3

35. Maximum magnification produced by simple micro-scope of focal length f = 5 cm is (1) 5 (2) 7 (3) 6 (4) Sol. Answer (3) Maximum magnification is when final image is at near point D. m

=

1+

D f

=

1+

25 5

8

[D = 25 cm]

=6 SECTION - C

Previous Years Questions

1.

In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eyepiece forms a real image of this line. The length of this image is I . The magnification [Re-AIPMT-2015] of the telescope is (1)

L

L

(2)

I

I

1

(3)

L I

–1

(4)

L

 I

L

– I

Sol. Answer (1)

Objective

eyepiece I

L v

f0 + f

e

f

0

At normal adjustment M =

...(i)

f

e

and distance between lenses = f 0 + f e Lateral magnification

L I



Using lens equation

f

0

 f

...(ii)

e

v

1

v

–

1

u



1

f







1

v 1 v



1

–



– f

0

f

0

 f

e





27

1

f

e

f



f


0

f

e



f

f

0

 f

0

e



 f

e

...(iii)

v

e

Comparing equations (i) (ii) & (iii) f

M =

0

f

e

2.



L I

A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47, respectively A

Blue Green Red

45° B

The prism will

C

[Re-AIPMT-2015]

(1) Separate the red colour part from the green and blue colours (2) Separate the blue colour part from the red and green colours (3) Separate all the three colours from one another (4) Not separate the three colours at all Sol. Answer (1) A

Red 45°

45° B

C

Refractive index of light rays that can just pass through the prism at grazing emergence at 2nd surface is

=

1 sin 45

= 1.414

Light having refractive index < 1.414 takes refraction but light having  > 1.414 suffers TIR.

 Only red colour light will come out of prism. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

28

3.



The refracting angle of a prism is A, and refractive index of the material of the prism is cot( A/2). The angle of [AIPMT-2015] minimum deviation is (1) 180° + 2 A

(2) 180° – 3 A

(3)

180° – 2 A

(4)

90° – A

Sol. Answer (3)

 A   m  2 ,

sin 



sin

cot

A

A

2

 A   m  2  

sin 



sin

2 90 

4.

A

A



2

2



A

2



cos sin

A

2 A

 A  m  2 

sin 



sin

2

A

2

   (180  2 A)

m

m

2

Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive [AIPMT-2015] index 1.7. The focal length of the combination is (1) 50 cm

(2) – 20 cm

(3)

– 25 cm

(4)

– 50 cm

Sol. Answer (4) 1

f

1 f1 1 f1

 

1 f



f2

1 f 2 1 f 2

1 40



f 

5.

f1



1

f 1

f 2

f3

 1 1    20  

 (1.5 1)  

1

f 3

40

 (1.7  1)  

f 3

f

1

1  0.7 7  1    10 100  20 20 

1

1

1

 



1 40



7 100

2 100

 50 cm

If the focal length of objective lens is increased then magnifying power of

[AIPMT-2014]

(1) Microscope will increase but that of telescope decrease (2) Microscope and telescope both will increase (3) Microscope and telescope both will decrease (4) Microscope will decrease but that of telescope will increase Sol. Answer (4)

MP of microscope 

L f0

 P  1    f  e

MP of telescope  f 0 1  f  e

fe



D



6.


The angle of a prism is A. One of its refracting surface is silvered. Light rays falling at an angle of incidence 2 A on the first surface returns back through the same path after suffering reflection at the silvered surface. The [AIPMT-2014] refractive index , of the prism is: (1) 2 sin A

(2) 2 cos A

(3)

1 2

cos A

Sol. Answer (2)

sin i sin r

2 A



sin2 A sin A

(4)

tan A

A

i = 2 A, r = A

7.

29

A

     2cos A

A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices 1 and 2 and R is the radius of curvature of the [NEET-2013] curved surface of the lenses, then the focal length of the combination is R

(1)

2

    1

2

(2)

R

 1  2 

(3)

2R

    2

1

(4)

R 2

    1

2

Sol. Answer (2)

8.

For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea [NEET-2013] eye lens can be estimated to be (1) 2.5 cm

(2) 1.67 cm

(3)

1.5 cm

(4)

5 cm

Sol. Answer (2)

9.

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the [AIPMT (Prelims)-2012] objective and eyepiece is 20 cm. The focal length of lenses are (1) 18 cm, 2 cm

(2) 11 cm, 9 cm

(3)

10 cm, 10 cm

(4)

15 cm, 5 cm

Sol. Answer (1) f o + f e = 20 f o f e

=

....(i)

9

....(ii)

Solving (i) and (ii) f o = 18 cm

f e = 2 cm

10. A ray of light is incident at an angle of incidence i on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. It the refractive index of the prism is , the angle of incidence i , is [AIPMT (Prelims)-2012] nearly equal to (1)

A



(2)

A 2

(3)

 A

(4)

 A 2


30



Sol. Answer (3) r 1 + r 2 = A

For ray to pass normally r 2 = 0

 r 1 = A µ=

sin i sin A

 sin A = sini If both A and i are small i = A 11. A concave mirror of focal length f 1 is placed at a distance of 'd' from a convex lens of focal length f 2 A beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance [AIPMT (Prelims)-2012] d must equal (1) 2f 1 + f 2

(2) –2f 1 + f 2

(3)

f 1 + f 2

(4)

–f 1 + f 2

Sol. Answer (1)

 d = 2f 1 + f 2 2f 1

f 2 d

12. When a biconvex lens of glass having refractive index 1.47 is dipped in a liqud, it acts as a plane sheet of [AIPMT (Prelims)-2012] glass. This implies that the liquid must have refractive index (1) Greater than that of glass

(2)

Less than that of glass

(3) Equal to that of glass

(4)

Less than one

Sol. Answer (3)

Power of lens is euqal to zero. 1 f

µ   1 1  =  g − 1  −  = 0  µe   R1 R 2 

 g = l 13. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a [AIPMT (Mains)-2012] material whose refractive index (1) Lies between (3) Is less than 1

2

and 1

(2)

Lies between 2 and

(4)

Is greater than 2

2

Sol. Answer (2)

 = i + e – A min, i = e min = A Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



31

2 A = 2i A = i

At min, sin A + δ m 2

=

= =

A

sin

2cos

=

2

sin A sin

A

2 sin

=

2

A

2

. cos

sin

A

2

A

2

A

2 i

2cos

( A = i )

2

i max = 90º

i min = 0º

 = 2 cos 45º

 = 2 cos 0º

=

 = 2

2

As R.I. lies between 2 and

2

.

14. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that [AIPMT (Mains)-2012] its end closer to the pole is 20 cm away from the mirror. The length of the image is (1) 10 cm

(2) 15 cm

(3) 2.5 cm

(4)

5 cm

Sol. Answer (4) O1 is at C , image of O1 will form at same postion.

For image of O2 u = –30 cm f = –10 cm 1

v

−

1

u

=

−30

f = 10 cm

O1

R = 20 cm

1

f

10 cm

1 +

v

1

O2

20 cm

P

1 =

−10

v = –15 cm

Length of image I 1I 2 = |PI 2 – PI 1|

= |15 – 20| = 5 cm 15. Which of the following is not due to total internal reflection?

[AIPMT (Prelims)-2011]

(1) Brilliance of diamond (2) Working if optical fibre (3) Difference between apparent and real depth of a pond (4) Mirage on hot summer days Sol. Answer (3)

Real & apparent depth are explained on the basis of refraction only. TIR not involved here. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

32



16. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describe best [AIPMT (Prelims)-2011] the image formed of an object of height 2 cm placed 30 cm from the lens? (1) Real, inverted, height = 1 cm

(2)

Virtual, upright, height = 1 cm

(3) Virtual, upright, height = 0.5 cm

(4)

Real, inverted, height = 4 cm

Sol. Answer (4) 1 f

1 1 = (1.5 − 1)  +   20 20 

f = 20 cm f

m=

f

+

u

20

m=

20 30

=

2 (Real and inversed)

−

−

hi h0

2 , h0

=

−

=

2 cm

i = –4 cm 17. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move [AIPMT (Mains)-2011] 5 cm closer to the lens. The focal length of the lens is (1) –30 cm

(2) 5 cm

(3) –10 cm

(4)

20 cm

Sol. Answer (1)

For virtual object u = +10 cm, v = +15 cm The ray diagram is as shown. 1

1

−

v

u

1 15

2

−

=

1

f

1 −

10

3

30

1 =

5 cm

f

1 =

15 cm

f

f = –30 cm

18. A thin prism of angle 15° made of glass of refractive index 1 = 1.5 is combined with another prism of glass of refractive index 2 = 1.75. The combination of the prisms produces dispersion without deviation. The angle of [AIPMT (Mains)-2011] the second prism should be (1) 12°

(2) 5°

(3)

7°

(4)

10°

Sol. Answer (4)

Angle = 15º

1 = 1.5 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



33

2 = 1.75 15(1 – 1) + A(2 – 1) = 0 7.5 + 0.75 A = 0 0.75 A = –7.5 7.5

A

=

−

0.75

 A = 10º 19. A ray of light travelling in a transparent medium of refractive index , falls on a surface separating the medium from air at an angle of incidence of 45°. For which of the following value of  the ray can undergo total internal [AIPMT (Prelims)-2010] reflection? (1)  = 1.25

(2)  = 1.33

(3)

 = 1.40

(4)

 = 1.50

Sol. Answer (4)

C < 45º sin C < sin 45º 1

< sin 45º

µ

>

2

Only possible with  = 1.5 20. A lens having focal length f and aperture of diameter d forms an image of intensity I . Aperture of diameter d / 2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be [AIPMT (Prelims)-2010] respectively (1) f and

I

(2)

4

3f 4

and

I 2

(3)

f and

3I 4

(4)

f 2

and

I 2

Sol. Answer (3)

Focal length will not change as long as curvature of lens does not change. I  d 2 d = Diameter of aperture I = Intensity of image d 2

= Aperture is covered by black paper 2

I 



I  =

d

4 I

4

obstructed by paper

Intensity of image =

I

−

I

4

=

3I 4


34



21. The speed of light in media M 1 and M 2 is 1.5 × 10 8 m/s and 2 × 10 8 m/s respectively. A ray of light enters from medium M 1 to M 2 at an incidence angle i . If the ray suffers total internal reflection, the value of i is [AIPMT (Mains)-2010]

2  

(1) Equal to sin–1   3 3  

(3) Equal to or greater than sin–1   4

3  

(2)

Equal to or less than sin–1   5

(4)

Less than sin–1   3

2  

Sol. Answer (3) i > C

× 108 sin i > 8 2 × 10 1.5

sin i > sin C

sin i >

sin i >

µ2

sin i >

µ1 v 1

i >

v 2

sin

3 4

−1

3   4 

22. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first [AIPMT (Mains)-2010] face (i.e., incident face) of the prism is (1) Zero

(2) 30°

(3)

45°

(4)

60°

Sol. Answer (2)

At minimum deviation r = r 

According to geometry of prism r + r  = A

2r = A r =

60 2

r = 30º

23. Two thin lenses of focal lengths f 1 and f 2 are in contact and coaxial. The power of the combinations is [AIPMT (Prelims)-2008]

(1)

f1  f 2 f1f 2

(2)

f 1 f 2

(3)

f 2 f 1

(4)

f1  f 2 2

Sol. Answer (1) 1

f

=

P =

1

+

f1 f1

+

1

f 2 f 2

f1f 2

1   f = P 




35

24. A boy is trying to start a fire by focusing Sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the Sun is 1.39 × 10 9 m and its mean distance from the earth is 1.5 × 10 11 m. What [AIPMT (Prelims)-2008] is the diameter of the Sun's image on the paper? (1) 12.4 × 10–4 m

(2) 9.2 × 10–4 m

(3)

6.5 × 10–4 m

(4)

6.5 × 10–5 m

Sol. Answer (2) v = 10 cm u = 1.5 × 10 11 m v

Magnification = 0.1 11

1.5 × 10 1

12

1.5 × 10

or

1.39 1.5

×

u

10 −3

Image diameter Sun's diameter

Image

=

×

=

1.39 × 10 9 1.39 × 109

=

=

Image

Image

or 9.2 × 10–4 = Image

 Image diameter = 9.2 × 10 –4 m 25. The frequency of a light wave in a material is 2 × 1014 Hz and wavelength is 5000 Å. The refractive index of [AIPMT (Prelims)-2007] material will be (1) 1.33

(2) 1.40

(3)

1.50

(4)

3.00

Sol. Answer (4) C

 =

v m C

=

(Medium) fm

f m λ m

= f a (air)

C

= =

f a λ m 3 × 10

8

14

2 × 10

×

5000 × 10

−10

=

3

26. A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid? [AIPMT (Prelims)-2007] 3 cm

4 cm

coin

(1) 1.2 × 108 m/s

(2) 1.8 × 108 m/s

(3)

1.3 × 108 m/s (4)

3.0 × 108 m/s


36



Sol. Answer (2) 3

 is critical angle sin =

5

=

1 µ l

5

 l =

3m

3

 µ l =

v in v in

4m

air (C )



medium

 v in medium

= =

3 5 9 5

5m

× 3 × 108 × 108

v = 1.8 × 10 8

27. A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus [AIPMT (Prelims)-2006] again? (1) 1 cm upward

(2) 4.5 cm downward

(3)

1 cm downward

(4)

2 cm upward

Sol. Answer (1)



1



Shift = t 1 − µ  =

 − 1   1.5 

3 1

= 1 cm So the microscope must be moved by 1 mm upwards µ=


Apparent depth =

Real depth µ

2 =

3

×

3 mm

= 2 mm

 Shift = 3 – 2 = 1 mm So the microscope must be moved 1 mm upwards. 28. A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a [AIPMT (Prelims)-2006] combination of lenses. The power in diopters of the combination is (1) 25

(2) 50

(3)

Infinite

(4)

Zero

Sol. Answer (4)

Each lens of same power but different is sign. When added P = P 1 + P 2 P = P 1 – P 2 P =0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



37

29. A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be (1) 12 feet

(2) 3 feet

(3)

6 feet

(4)

Any length

Sol. Answer (2)

Height - 6 ft To see any object in a plane mirror complete, a mirror must be half the height of object. So minimum height of mirror =

h 2

= 3 ft

30. Images formed by an object placed between two plane mirrors whose reflecting surfaces make an angle of 90° with one another lie on a (1) Straight line

(2) Parabola

(3)

Circle

(4)

Ellipse

Sol. Answer (3) m2

I2

O

m1

I3

I1

31. An object is placed between two plane mirrors inclined at an angle ‘ ’ to each other. If the number of images formed is 7, then the angle of inclination ‘ ’ is (1) 15°

(2) 30°

(3)

45°

(4)

60°

(1) Greater than that of the cladding

(2)

Equal to that of the cladding

(3) Smaller than that of the cladding

(4)

Independent of that of the cladding

Sol. Answer (3)

7=

360

−1

θ

 = 45º 32. In optical fibres, the refractive index of the core is

Sol. Answer (1)

33. Light travels through a glass plate of thickness t and having a refractive index . If c is the velocity of light in vacuum, the time taken by light to travel this thickness of glass is t

(1)  c

(2)

 t c

(3)

t  c

(4)

tc



Sol. Answer (2)

Time =

= V =

t v (Speed

in glass)

c V c

µ

Time =

µt

c


38



34. A ray of light from a denser medium strikes a rare medium as shown in figure. The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r . The critical angle would be A

N

C

i r DENSER RARER B r  D N 

(1) sin–1 (tan r )

(2) tan–1 (sin r )

(3)

sin–1 (cot r )

(4)

tan–1 (sin r )

(4)

2.25 × 108 m/s

Sol. Answer (1) i = r r + r  = 90º i + r  = 90º r  = 90º – i

sin c =

1 µ (R.I. of denser w.r.t. rarer)

sin c =  (R.I. of rarer w.r.t. denser) sin c = sin c =

sin i ′

sin r

sin i sin(90º −i )

sin c = tan i

(i = r )

c = sin–1 (tan r)

35. The refractive index of water is 1.33. What will be the speed of light in water? (1) 4 × 108 m/s

(2) 1.33 × 108 m/s

(3)

3 × 108 m/s

Sol. Answer (4)

1.33 =

c V (Speed

of light in water)

36. An electromagnetic radiation of frequency n, wavelength l, travelling with velocity v in air, enters a glass slab of refractive index . The frequency, wavelength and velocity of light in the glass slab will be respectively (1) n, 2 and

v



(2)

   and v

2n

,

(3)

v  and    n

,

(4)

n,

v  and  

Sol. Answer (4)

By definition of refractive index Velocity of light becomes

v

µ

λ

 becomes µ But frequency is constant. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



39

37. A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out is (1) 

(2) 3 m

(3)

6m

(4)

4m

Sol. Answer (2) i c =

 r

sin i C =

sin i C =

1

i C

µ d

4m



3 5

3

 tan i c =

25

− 32

=

3 4

r = 4 tan

 r = 3 m 38. A ray of light travelling in air have wavelength , frequency n, velocity v and intensity I . If this ray enters into water then these parameters are , n, v  and I  respectively. Which relation is correct from following? (1)  = 

(2) n = n

(3)

v = v 

(1) Total internal reflection

(2)

Less scattering

(3) Refraction

(4)

Less absorption coefficient

(4)

I = I 

Sol. Answer (2)

When ray enters water λ

v

 becomes µ and velocity becomes µ Intensity also changes Only frequency n remains same so answer is (2). 39. Optical fibre are based on

Sol. Answer (1)

Optical fibres depend on total internal reflection. 40. For the given incident ray as shown in figure, the condition of total internal reflection of this ray the required refractive index of prism will be 45º

(1)

3  1 2

(2)

2  1 2

incident ray

(3)

3 2

(4)

7 6


40



Sol. Answer (3) i C = 90 – r

45º



   µg 

sin(90 − r ) = 

Also



1

sin45º µg =

i C

r

....(ii)

sin r

sin r

cos r

=

sin45º

tanr = sin45º =

sin r

....(i)

1 2

 1

1 =

3

r

2 

sin i

µ =

sin r

3 =

2

41. A layer of benzene ( = 1.5) 12 cm thick floats on water layer (m = 4/3) 8 cm thick in a vessel. When viewed from the top, the apparent depth of bottom of vessel below the surface of benzene will be (1) 20 cm

(2) 14 cm

(3)

7 cm

(4)

21 cm

Sol. Answer (2)

In case of multiple medium of different R.I. Apparent depth

d

12 =

d =

t1

µ1

+

t 2

µ2

8

1.5

+

4/3

d=8+6

= 14 cm 42. The critical angle for a light travelling from medium A into medium B is . The speed of light in medium A is v , the speed of light in medium B is (1)

v

cos



(2) v sin 

(3)

v

sin 

(4)

v cos



Sol. Answer (3) sin θ

=

µB µ A

 A sin = B Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


Ray Optics and Optical Instruments C

C µ A =

C v



µB =

v

v B

C

sin θ =

v B

v

=

v B

41

sin θ

43. A small bulb is placed at a depth of 2 7 m in water and floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. The minimum diameter of the disc is ( water = 4/3) (1) 42 m

(2) 6 m

(3)

2 7

m

(4)

12 m

Sol. Answer (4) r

θmax = sin−1

1

µ

= sin −1

3



4 27

3

sin  =



4 3

 tan  =

7

3

r =

2 7

7

r = 6 m

44. Two optical media of refractive indices 1 and 2 contain x and y number of waves in the same thickness. Their  relative refractive index 2 is equal to 1 (1) xy

y

(2)

x

(3)

x y

(4)

y  x x

Sol. Answer (2)

let wavelength of light in air is  λ

 Wavelength in medium = µ t =

x

µ1

=

y



x λ

µ2 µ1

and

t =

y λ

µ2

µ1 µ2

=

y x


42



45. A light ray is travelling through a ring of an optical fibre which is made of four different glasses (shown below) but each part has the same geometrical thickness. Their respective refractive indices are shown. The light ray will take the maximum time in crossing the part Part-I

Part-II

1.51 1.52 1.54 1.53 Part-IV

(1) I

(2) II

Part-III

(3)

IV

(4)

Same in all

Sol. Answer (3)

R.I. of part IV is highest. So velocity will be lowest. Hence, maximum time is taken in part IV. 46. Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever be the value of angle of incidence? (1) n = 1.1

(2) n = 1

(3)

n



2

(4)

n = 1.3

Sol. Answer (3)

sin < n>

n

2

−1



n

2

sin θ + 1

Maximum value of  = 90º. 2

n >

47. If the refractive index of a material of equilateral prism is is (1) 60°

(2) 45°

(3)

3

, then angle of minimum deviation of the prism

30°

(4)

75°

Sol. Answer (1)

R.I. =

; A = 60º; m = ?

3

( +δ )

sin A

m

2

µ= sin

 A   2 

sin(60º + δm ) 2 sin30º

3=

−1 sin

3

=

60 + δ m

2

60º = 30º +

30º =

2

δm 2

δm 2

m = 60º Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



43

48. A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass slab. When coming out on the opposite parallel face, the red and green ray emerge from (1) Two points propagating in two different non parallel directions (2) Two points propagating in two parallel directions (3) One point propagating in two different directions (4) One point propagating in the same directions Sol. Answer (2)

Since parallel beams of light will regain their original direction. They will again become parallel after emergence. 49. The refractive index of the material of a prism is 2 and its refracting angle is 30°. One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is (1) 45°

(2) 60°

(3)

Zero

(4)

30°

Sol. Answer (1)

By geometry r = 90º – 60º = 30º 30°

3



=

sin i

sin i sin30º

i

r

1 =

2

i = 45º

50. Four lenses of focal length ±15 cm and ±150 cm are available for making a telescope. To produce the largest magnification, the focal length of the eyepiece should be (1) +15 cm

(2) +50 cm

(3)

–150 cm

(4)

–15 cm

Sol. Answer (1)

51. A lens is placed between a source of light and a wall. It forms images of area A1 and A2 on the wall, for its two different positions. The area of the source of light is (1)

A1

 A2

(2)

2

1 A1



1 A2

(3)

A1 A2

(4)

A1

 A2 2

Sol. Answer (3)

This is application of the displacement method for finding focal length. Here m1m2 = 1 Let A0 be area of object A1 A0

A0

×

=

A2 A0

=

1

A1A2


44



52. If f V and f R are the focal lengths of a convex lens for violet and red light respectively and F V and F R are the focal lengths of a concave lens for violet and red light respectively, then we must have (1) f V > f R and F V > F R

(2) f V < f R and F V > F R

(3)

f V > f R and F V < F R (4)

f V < f R and F V < F R

Sol. Answer (2)

v > r A converging lens with higher refractive index will converge rays more hence value of f v < f r Same is true for concave lenses but since values for focal length are taken as negative F v > F r 53. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together, what will be their resulting power? (1) + 7.5 D

(2) – 0.75 D

(3)

+ 6.5 D

(4)

– 6.5 D

Sol. Answer (2) P = P 1 + P 2 1 =

1

0.8

5 =

−

4

−

0.5

2

= –0.75 D 54. The focal length of a converging lens is measured for violet, green and red colours. It is respectively f v , f g, f r . We will get (1) f v < f r

(2) f g > f r

(3)

f v = f g

(4)

f g = f r

Sol. Answer (1)

v > g > r A converging lens with greater refractive index will bend rays more converging them closer.

 f v < f g < f r 55. A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order to have an upright image of the object coincident with it? (1) 50 cm

(2) 30 cm

(3) 12 cm

(4)

60 cm

Sol. Answer (1) 20 cm

For convex lens 1 +20

1 =

1 −

v

−30

O

C d = 60 – 10

v = 60 cm

Virtual object for convex mirror should be at its C .

d = 50 cm

30 cm

d

10 cm 60 cm




45

56. A plano-convex lens is made of refractive index 1.6. The radius of curvature of the curved surface is 60 cm. The focal length of the lens is (1) 200 cm

(2) 100 cm

(3)

50 cm

(4)

400 cm

Sol. Answer (2) 1 f

1 1 = ( µ − 1)  −   R ∞ 

1 f

0.6 =

60

f = 100 cm

57. A planoconvex lens (m = 1.5) has radius of curvature 10 cm. It is silvered on its plane surface. Find focal length after silvering (1) 10 cm (2) 20 cm (3) 15 cm (4) 25 cm Sol. Answer (1)

 = 1.5 R = 10 cm P = 2P L + P m P m = Power of mirror = 0  P = 2P L 1 1 = (1.5 − 1)  +   10 ∞ 

P L

0.5 =

1

P L

=

10

cm

20

P L = 5 D

 P = 2P L = 10 D f

1 =

10

m =

10 cm

58. A bubble in glass slab (m = 1.5) when viewed from one side appears at 5 cm and 2 cm from other side, then thickness of slab is (1) 3.75 cm

(2) 3 cm

(3)

10.5 cm

(4)

2.5 cm

Sol. Answer (3)

 = 1.5 µ=


Real depth = Apparent depth 1 t = t 1 + t 2

t 1 t t 2

t = 5 × 1.5 + 2 × 1.5

= 10.5 cm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

46



59. A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of a convex lens. If the distance between the two walls is d , then required focal length will be (1) Only

d 4

(3) More than

d 4

but less than

d 2

d

(2)

Only

(4)

Less than or equal to

2

d 4

Sol. Answer (4) f 

d 4

60. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will (1) Become zero

(2)

Become infinite

(3) Become small, but non-zero

(4)

Remain unchanged

Sol. Answer (2)

When this happens the lens will behave as a glass slab and parallel rays will converge at infinity.

 f =  61. An equiconvex lens is cut into two halves along (i) XOX  and (ii) YOY  as shown in the figure. Let f , f , f  be the focal lengths of the complete lens of each half in case (i), and of each half in case (ii), respectively. Choose the correct statement from the following Y

X 

X

O

Y 

(1) f  = f , f  = 2f

(2) f  = 2f , f  = f

(3)

f  = f , f  = f

f  = 2f , f  = 2f

(4)

Sol. Answer (1) f = f  as radius of curvature of both surfaces is same. Y

f  = 2f

by lens maker's formula : 1 1 = (µ − 1)  −   R ∞  f ''

....(i)

1 1 = (µ − 1)  +   R R 

....(ii)

1

1 f f

f ''

=

1 2

X 

O

X

Y 

[Dividing (i) and (ii)]

 f  = 2f Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



47

62. Which of the following is incorrect? (1) A thin convex lens of focal length f 1 is placed in contact with a thin concave lens of focal length f 2. The combination will act as convex lens if f 1 < f 2 (2) Light on reflection at water-glass boundary will undergo a phase change of  (3) Spherical aberration is minimized by achromatic lens (4) If the image of distant object is formed in front of the retina then defect of vision may be myopia Sol. Answer (3)

Achromatic lenses minimize chromatic aberration. 63. A double concave thin lens made out of glass (  = 1.5) have radii of curvature 500 cm. This lens is used to rectify the defect in vision of a person. The far point of the person will be at (1) 5 m

(2) 2.5 m

(3) 1.25 m

(4)

1m

Sol. Answer (1) 1

f

=

1 v

−

1 ∞

 

(1.5 − 1)  −

=1  500  v 2

v = – 500 cm v = – 5 m

64. A convex lens forms a real image 16 mm long on a screen. If the lens is shifted to a new position without disturbing the object or the screen then again a real image of length 81 mm is formed. The length of the object must be (1) 48.5 mm

(2) 36 mm

(3)

6 mm

(4)

72 mm

Sol. Answer (2)

This is the application of the displacement method. Size of object (m0) =



m0 =

m m 1 2

16 × 81

=4×9 = 36 cm 65. A point object is moving with speed u 0 at a position somewhere between 2 F and F in front of a convex lens. The speed of its image is u 0 2F

(1) > u 0

(2) < u 0

F

F

(3)

2F

= u 0

(4)

May be (1) or (2)


48



Sol. Answer (1) 1

v

1

−

1

=

u

f

Differentiating w.r.t. t

−

1 du −  − 2  =0  u  dt

1 dv v

2

dt

1 − v

1 v

2

i

2 v

vi =

2 u

+

2 u

v 0 =

0

v0

When f < u < 2f , v lies beyond 2 f .  v > 2f v

2

2 u

>

1

v i > v 0.

66. The minimum magnifying power of a telescope is M . If focal length of its eye lens is halved, the magnifying power will become M

(1)

(2) 2M

2

(3)

3M

(4)

4M

Sol. Answer (2) M =

f 0 f e

M 

f e =

1

f e f e 2

M  = 2M

67. An object is placed in front of two convex lenses one by one at a distance u from the lens. The focal lengths of the lenses are 30 cm and 15 cm respectively. If the size of image formed in the two cases is same, then u is (1) 15 cm

(2) 20 cm

(3)

25 cm

(4)

30 cm

Sol. Answer (2)

Magnification are same

 m = f    f + u 

m1 = – m2 30 30

15 =

−

u

−

15

−

u

 u = 20 cm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



49

68. If R 1 and R 2 are the radii of curvature of the spherical surfaces of a thin lens and R 1 > R 2, then this lens can R 2 R 1

(1) Correct myopia

(2)

Correct hypermetropia

(3) Correct presbiopia

(4)

Correct astigmatism

Sol. Answer (1) f < 0 (Diverging)

69. The focal length of a thin lens in vacuum is f . If the material of the lens has   immersed in water of refractive index (1) f

(2)

4 3

4 3

3 2

, its focal length when

will be

f

(3)

2f

(4)

4 f

Sol. Answer (4)

 3 − 1  1 − 1  =  2    R1 R 2 

1

f v 1

f

1 1 = 2  R − R  1 2

1

f w 1

f w

1

 µg



9

 2

1

1



=  µ − 1  R − R   w  1 2 =  − 1   8 f

 f w = 4f 70. For a telescope having f o as the focal length of the objective and f e as the focal length of the eyepiece, the length of the telescope tube is (1) f e

(2) f o – f e

(3)

f o

(4)

f o + f e

Sol. Answer (4)

By the construction of the telescope f 0 + f e = L 71. An astronomical telescope of tenfold angular magnification has a length of 44 cm. The focal length of the objective is (1) 44 cm

(2) 440 cm

(3)

4 cm

(4)

40 cm

Sol. Answer (4) m = 10 f 0 f e

=

10

f 0 + f e = 44 cm

....(i) ....(ii)

Solving (i) and (ii) f 0 = 40 cm Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

50



72. Ray optics is valid, when characteristic dimensions are (1) Much smaller than the wavelength of light

(2)

Of the same order as the wavelength of light

(3) Of the order of one millimeter

(4)

Much larger than the wavelength of light

Sol. Answer (4)

Ray optics is valid on a more macro scale compared to wavelength of light. On micro scale wave optics and wave-particle duality is more prominent. 73. The blue colour of the sky is due to the phenomenon of (1) Scattering

(2) Dispersion

(3)

Reflection

(4)

Refraction

Sol. Answer (1)

Blue colour of the sky is due to scattering of light also called the Rayleigh scattering. 74. Rainbow is formed due to (1) Scattering and refraction

(2)

Total internal reflection and dispersion

(3) Reflection only

(4)

Refraction and dispersion

Sol. Answer (4)

Rainbow's colours appear due to dispersion, which occurs after refraction. SECTION - D

Assertion - Reason Type Questions

1.

A : Plane and convex mirrors can produce real images of objects. R : A plane or convex mirror can produce a real image if the object is virtual.

Sol. Answer (1)

Real images are formed from real intersection of light rays, which occur in convex mirror when the object is virtual. 2.

A : A virtual image cannot be caught on a screen but it can be photographed. R : The virtual image here serves as an object for the lens of the camera to produce a real image.

Sol. Answer (1)

Virtual image can be observed but real intersection of light rays needs to occur to take objects to screen. 3.

A : When a diver under water looks obliquely at a fisherman standing on the bank of a lake then fisherman looks taller. R : When a ray of light travelling in air enters water, it bends towards the normal.

Sol. Answer (1)

The ray of light bending towards normal makes it appear that light comes from further away. 4.

A : The apparent depth of a tank of water decreases if viewed obliquely. R : Real depth decreases if viewed obliquely.

Sol. Answer (3)

Apparent depth of tank decreases because the light ray, travels further through the water bending more in the process. Real depth remains the same are it was. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


5.


51

A : Proper cutting of diamond makes it sparkle. R : Diamond has very large refractive index.

Sol. Answer (1)

Proper cutting of diamond makes it sprakle due to total internal refraction. A high reflective index makes total internal reflection easier. 6.

A : The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass so the magnification produced is one. R : During image formation through magnifying glass, the object as well as its image are at the same position.

Sol. Answer (4)

A magnifying glass magnifies the virtual image and magnification is greater than one. The reason is also false. 7.

A : In viewing through a magnifying glass, angular magnification decreases if the eye is moved back. R : Angle subtended at the eye becomes slightly less than the angle subtended at the lens.

Sol. Answer (1)

8.

A : Magnifying power of a simple microscope cannot be increased beyond a limit. R : Magnifying power is inversely proportional to focal length and there are some practical difficulty of grinding, aberrations etc. because of which focal length cannot be decreased below a limit.

Sol. Answer (1)

There is a limit to how curved a glass piece can be made practically. 9.

A : The objective and the eye-piece of a compound microscope should have short focal lengths. R : Magnifying power of a compound microscope is inversely proportional to the focal lengths of both the lenses.

Sol. Answer (1)

For normal adjustment : M =

LD fo ⋅ f e

.

10. A : When viewing through a compound microscope, our eyes should be positioned not on the eye piece but a short distance away from it for best viewing. R : The image of the objective in the eye-piece is known as 'eye-ring' and if we position our eyes on the eyering and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. Sol. Answer (1)

11. A : The peculiar fish Anableps anableps swims with its eyes partially extending above the water surface so that it can see simultaneously above and below water. R : This fish has egg shaped eye lens and two retina. Sol. Answer (1)

12. A : A virtual image cannot be produced on a screen. R : The light energy does not meet at the point(s) where virtual image is formed. Sol. Answer (1)

To take a sharp image on a screen light rays must intersect really. This does not happen in virtual images. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Sumanta Chowdhury - CLS Aipmt-15-16 XIII Phy Study-Package-7 Set-1 Chapter-24

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