Chapter # 17
1. Sol.
Light Waves
SOLVED EXAMPLES
The refractive index of glass is 1.5. Find the speed of light in glass. We have
speed of light in vacuum = speed of light in the material Thus, speed of light in glass
2.
Sol.
=
speed of light in vacuum
=
30 10 8 m / s = 20 × 108 m/s. 1.5
In a Young’s double slit experiment, the separation between the slits in 010 mm, the wavelength of light used is 600 nm and the interference pattern in observed on a screen 1.0 m away. Find the separation between the successive bright fringes. The separation between the successive bright fringes is w=
3. Sol.
The wavelength of light coming from a sodium source is 589 nm. What will be its wavelength in water? Refractive index of water = 1.33. The wavelength in water is = 0/m, where 0 is the wavelength in vacuum and m is the refractive index of water. Thus, =
4. Sol.
Sol.
589 = 443 nm. 1.33
Find the minimum thickness of a film which will strongly reflect the light of wavelength 589 nm. The refractive index of the material of the film is 125. For strong reflection, the least optical path difference introduced by the film should be /2. The optical path difference between the waves reflected from the two surfaces of the film is 2d. Thus, for strong reflection. 2d = /2 or,
5.
D 10 m 600 10 9 m = 60 × 10–3 m = 6.0 mm. d 0.10 10 3 m
d
589 nm = = 118 nm. 4 1.25 4
A parallel beam of monochromatic light of wavelength 480 nm passes through a long slit of width 0.2 mm. Find the angular divergence in which most of the light is diffracted. Most of the light is diffracted between the two first order minima. These minima occur at angles given by b sin = ± or,
sin = ± /b
=±
450 10 2 m 0.2 10 2
= ± 2.25 × 10–3 .
or, = ± 2.25 × 10–3 rad. The angular divergence = 4.5 × 10–3 rad. 6. Sol.
A beam of light of move length 590 nm is focussed by a converging lens of diameter 10.0 cm at a distance of 20 cm from it. Find the diameter of the disc image formed. The angular radius of the central bright disc in a diffraction pattern from circular aperture is given by sin =
1.22 b
1.22 590 10 9 m
= 1.4 × 10–5 rad. 5.0 10 7 m The radius of the bright disc is 1.4 × 10–5 × 20 cm = 2.8 × 10–4 cm. The diameter of the disc image = 5.6 × 10–4 cm. =
manishkumarphysics.in
Page # 1
Chapter # 17
QUESTIONS
Light Waves
FOR
SHORT
ANSWER
1.
Is the colour of 620 nm light and 780 nm light same? Is the colour of 620 nm light and 0.21 light same? How many colours are there is white light?
2.
The wavelength of light in a medium is = 0/, where (0 = 720 nm) enters into water. The wavelength in water. The wavelength in water is = 0/ = 540 nm. To a person under water does this light appear green?
3.
Whether the diffraction effects from a slit will be more clearly visible or less clearly, if the slit-width is increased?
4.
It we put a cardboard (say 20 cm × 20 cm) between a light source and our eyes, we can’t see the light. But when we put the same cardboard between a sound source and our ear, we hear the sound almost clearly. Explain.
5.
TV signals broadcast by Delhi studio cannot be directly received at Patna which is about 1000 km away. But the same signal goes some 36000 km away to a satellite, gets reflected and is then received at patna. Explain.
6.
Can we perform Young’s double slit experiment with sound waves? To get a reasonable “fringe pattern”, wheat would be the order of separation between the slits? How can the bright fringes and the dark fringes to detected in this case?
7.
Is it necessary to have two waves of equal intensity to study interference patter? Will there be an effect on clearly if the waves have unequal intensity?
8.
Can we conclude from the interference phenomenon whether light is a transverse wave or a longitudinal wave?
9.
Why don’t we have interference when two candles are placed close to each other and the intensity is seen at a distant screen? What happens if the candles are replaced by laser sources?
10.
If the separation between the slits in a Young’s double slit experiment is increased, what happens to the fringes-width? If the separation is increased too much, will the fringe pattern remain detectable?
11.
Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing = 400 nm). Describe the nature of the fringe pattern observed.
Objective - I 1.
2.
Light is (A) wave phenomenon (B) particle phenomenon (C*) both particle and wave phenomenon izdk'k gS (A) rjax izÑfr dk (B) d.k dh izÑfr dk (C*) d.k
o rjax nksuksa izÑfr dk
The speed of light depends (A) on elasticity of the medium only (C) on elasticity as well as inertia
(B) on inertia of the medium only (D*) neither on elasticity nor on inertia
izdk'k dh xfr fuHkZj djrh gSA (A) dsoy ek/;e dh izR;kLFkk ij (C) tM+Ro rFkk izR;kLFkk ij
(B) dsoy ek/;e ds (D*) ;k rks izR;kLFkk
tM+Ro ij ij uk gh tM+Ro ij
3.
The equation of a light wave is written as y=A sin(kx-t). Here, y represents (A) displacement of either particles (B) pressure in the medium (C) density of the medium (D*) electric field izdk'k dh rjax dk lehdj.k y = A sin(kx –t) gS] ;gk¡ x iznf'kZr djrk gS (A) bZFkj ds d.kksa dk foLFkkiu (B) ek/;e esa nkc (C) ek/;e dk ?kuRo (D*) fo|qr {ks=k
4.
Which of the following properties show that light is a transverse wave? (A) Reflection (B) Interference (C) Diffraction fuEu esa ls dkSulh izÑfr iznf'kZr djrh gS fd izdk'k vuqizLFk rjaxsa gS (A) ijkorZu (B) O;frdj.k (C) foorZu manishkumarphysics.in
(D*) Polarization (D*) /kzo q .k Page # 2
Chapter # 17
Light Waves
5.
When light is refracted into a medium, (A) its wavelength and frequency both increase (B) its wavelength increases but frequency remains unchanged (C*) its wavelength decreases but frequency remains unchanged (D) its wavelength and frequency both decrease tc izdk'k ek/;e esa viofrZr gksrk gS rks (A) bldh rjaxnS/;Z rFkk vko`fÙk nksuksa esa o`f) gksrh gSA (B) bldh rjaxnS/;Z esa o`f) gksrh gS] ijUrq vko`fÙk vifjofrZr jgrh gSA (C*) bldh rjaxnS/;Z esa deh gksrh gS] ijUrq vko`fÙk vifjofrZr jgrh gSA (D) bldh rjaxnS/;Z o vko`fÙk nksuksa esa deh gksrh gSA
6.
When light is reflected, which of the following does not change? (A) Wavelength (B*) Frequency (C) Velocity tc izdk'k vifjofrZr gksrk gS rks fuEu esa ls D;k ifjofrZr ugha gksrk gS (A) rjaxnS/;Z (B*) vko`fÙk (C) osx
(D) Amplitude (D) vk;ke
7.
The amplitude modulated (AM) radio wave bends appreciably round the corners of a 1 m x 1 m board but the frequency modulated (FM) wave only negligibly bends. If the average wavelength of AM and FM waves are a and f, (A*) a > f (B) a = f (C) a < f (D) we don’t have sufficient information to decide about the relation of a and f vk;ke eksM;~ y w Vs M s (AM) jsfM;ks rjaxas 1eh- x 1eh- ds cksMZ ds fdukjksa ls eqM+ tkrh gS] ijUrq vko`fÙk eksM;~ y w Vs M s (FM) rjaxas ux.; eqM+rh gSA ;fn AM rFkk FM rjaxksa dh vkSlr rjaxnS/;Z a o f gks rks (A*) a > f (B) a = f (C) a < f (D) a o f esa lac/a k Kkr djus ds fy, lwpuk,¡ vi;kZIr gSA
8.
Which of the following sources gives best monochromatic light? (A) A candle (B) A bulb (C) A mercury tube fuEu esa ls fdl lzkrs ls loksÙZ ke ,do.khZ; izdk'k izkIr gksxk (A) eksecÙkh (B) cYc (C) edZfj V~;wc
(D*) A laser (D*)
ystj
9.
The wavefronts of light wave travelling in vacuum are given by x + y + z = c. The angle made by the direction of propagation of light with the X-axis is [HCV_Obj-1_Q. 9] fuokZr esa xfr dj jgh izdk'k rjax dk rjxkaxz x + y + z = c gSA izdk'k dh xfr dh fn'kk dk x-v{k ls cuk;k x;k dks.k gksxk (A) 0o (B) 45o (C) 90o (D*) cos-1 (1/3)
10.
The wavefronts of light coming from a distant source of unknown shape are nearly (A*) plane (B) elliptical (C) cylindrical (D) spherical vKkr vkÑfr ds nwj fLFkr lzkrs ls vkus okys izdk'k dk rjxkaxz yxHkx gksxk (A*) lery (B) nh?kZo`Ùkkdkj (C) csyukdkj (D) xksykdkj
11.
The inverse square law of intensity (i.e., the intensity (A*) point source
(B) line source
1 ) is valid for a r2
(C) plane source
1 ) ds O;qRØe oxZ dk fu;e ykxw gksrk gS r2 (A*) fcUnq lzksr ds fy, (B) js[kh; lzksr ds fy, (C) lery lzksr ds fy,
(D) cylindrical source
rhozrk ¼rhozrk
(D) csyukdkj
lzksr ds fy,
12.
Two source are called coherent if they produce waves (A) of equal wavelength (B) of equal velocity (C) having same shape of wavefront (D*) having a constant phase difference nks lzksr dyk lEc) dgykrs gSa] ;fn os rjaxsa mRiUu djrs gSa (A) leku rjaxnS/;Z dh (B) leku osx dh (C) leku vkÑfr ds rjxkaxz dh (D*) vpj dykUrj dh
13.
When a drop of oil is spread on a water surface, it displays beautiful colours in daylight because of (A) dispersion of light (B) reflection of light (C) polarization of light (D*) interference of light
tc rsy dh cwna dks ty dh lrg ij QSyk;k tkrk gS rks ;g fnu ds izdk'k esa lqUnj jaxksa dk fuEu ?kVuk ds dkj.k izn'kZu manishkumarphysics.in
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Chapter # 17
Light Waves
djrh gS (A) izdk'k ds o.kZ fo{ksi.k (C) izdk'k dk /kzo q .k 14.
(B) izdk'k dk ijkorZu (D*) izdk'k dk O;frdj.k
Two coherent sources of different intensities send waves which interfere. The ratio of maximum to minimum intensity is 25. The intensities of the sources are in the ratio -
nks dyk lEc) lzkrs ksa ls NksM+h xbZ vyx&vyx rhozrk dh rjaxas v/;kjksfir gksrh gSA vf/kdre rhozrk o U;wure rhozrk esa vuqikr 25 gS rks lzkrs ksa dh rhozrk esa vuqikr gksxk (A) 25: 1 15.
(B) 5: 1
(C*) 9: 4
(D) 625: 1
The slits in a Young’s double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is Io. If one of the slits is closed, the intensity at this point will be
;ax ds f}js[kkfNnz iz;ksx esa js[kkfNnzkas dh pkSM+kbZ leku gS rks lzkrs dks js[kkfNnzkas ds lkis{k lEfer j[kk tkrk gSA dsUnzh; fÝat dh rhozrk 0 gS] ;fn ,d js[kkfNnz dks
17.
(A) I0 (B*) IO/4 (C) I0/2 (D) 4I0 A thin transparent sheet is placed in front of a Young’s double slit. The fringe-width will (A) increase (B) decrease (C*) remain same (D) become nonuniform ;ax ds f}&js[kkfNnz iz;ksx esa ,d iryh ikjn'kZd ifV~Vdk j[k nh tkrh gS rks fÝat pkSM+kbZ (A) c<+sxh (B) ?kVsxh (C*) leku jgsxh (D) vleku gksxh If Young’s double slit experiment is performed in water (A*) the fringe width will decrease (B) the fringe width will increase (C) the fringe width will remain unchanged (D) there will be no fringe ;fn ;ax ds f}js[kkfNnz iz;ksx dks ikuh esa fd;k tk;s rks (A*) fÝat pkSM+kbZ de gks tk;sxhA (B) fÝat pkSM+kbZ c<+ tk;sxhA (C) fÝat pkSM+kbZ vifjofrZr jgsxhA (D) dksbZ fÝat izkIr ugha gksxhA
Objective - II 1.
A light wave can travel (A*) in vacuum (C*) in a material medium
(B) in vacuum only (D) in a material medium only
izdk'k rjax xfr dj ldrh gS (A*) fuokZr~ esa (C*) inkFkZ ek/;e esa
(B) dsoy (D) dsoy
fuokZr esa inkFkZ ek/;e esa
2.
Which of the following properties of light conclusively support wave theory of light (A) Light obeys laws of reflection (B*) Speed of light in water is smaller than the speed in vacuum (C*) Light shows interference (D) Light shows photoelectric effect. fuEu esa ls izdk'k dh dkSulh izÑfr izdk'k dh rjax izÑfr ds i{k esa gS (A) izdk'k ijkorZuds fu;e dk ikyu djrk gSA (B*) ty esa izdk'k dk osx fuokZr esa osx ls de gSA (C*) izdk'k ds lapj.k dh fn'kk ds yEcor~ gSA (D) izdk'k] izdk'k fo|qr izHkko iznf'kZr djrk gSA
3.
When light propagates in vacuum there is an electric field and a magnetic field. These fields (A) are constant in time (B*) have zero average value (C*) are perpendicular to the direction of propagation of light (D*) are mutually perpendicular tc izdk'k fuokZr esa xeu djrk gS rks mlesa fo|qr o pqEcdh; {ks=k gksrs gSAa ;s {ks=k (A) le; ds lkFk vpj gSA (B*) 'kwU; vkSlr eku okys gSAa (C*) izdk'k ds lapj.k dh fn'kk ds yEcor~ gSA (D*) ,d nwljs ds yEcor~ gSA manishkumarphysics.in
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Chapter # 17 Light Waves 4. Huygen’s principle of secondary wavelets may be used to (A) find the velocity of light in vacuum (B) explain the particle behavior of light (C*) find the new position of a wavefront (D*) explain Snell’s law gkbxsu ds f}rh;d rjafxdk fl)kUr dk mi;ksx fd;k tkrk gS (A) fuokZr esa izdk'k dk osx Kkr djus esa (B) izdk'k dh rjax izÑfr dks crkus esaA (C*) rjxkaxz dh ubZ fLFkfr Kkr djus esaA (D*) Lusy ds fu;e dks le>kus esaA 5.
6.
Three observers A, B and C messure the speed of light coming from a source to A, B and C. The observer A movies towards the source and C moves away from the source at the same speed. The observe B stays stationery. The surrounding space is vaccum everywhere. rhu ijh{kdksa A, B rFkk C }kjk lzkrs ls fuxZr izdk'k dh ekih xbZ xfr A, B vkSj C gSA ijh{kd A lzkrs dh rjQ rFkk ijh{kd C lzksr ls nwj leku xfr ls pyrk gSA ijh{kd B fLFkj jgrk gSA ek/;e fuokZr gks rks 1 (A) A> B > C (B) A< B < C (C*) A= B = C (D*) B= A + C) 2 Suppose the medium in the previous question is water. Select the correct option(s) from the list given in that question. ekuk fiNys iz'u esa ek/;e ty gS rks lgh fodYi pqfu;s (Ans = A, D)
7.
Light waves travel in vacuum along the X-axis. Which of the following may represent the wavefronts? izdk'k fdj.ksa fuokZr~ esa x-v{k ds vuqfn'k xfr dj jgh gSA fuEu esa ls rjxkaxz dks iznf'kZr djrk gS (A*) x = c (B) y = c (C) z = c (D) x + y + z = c
8.
If the source of light used in a Young’s double slit experiment is changed from red to violet, (A) the fringes will become brighter (B*) consecutive fringes will comes closer (C) the intensity of minima will increase (D) the central bright fringe will become a dark fringe ;fn ;ax ds f}js[kkfNnz iz;ksx esa izdk'k ds yky ls cSaxuh esa ifjofrZr dj nsa rks (A) fÝat vf/kd pedhyh gks tk;sxh (B*) Øekxr fÝats vkSj vf/kd ikl vk;sxh (C) fufEu"B dh rhozrk c<+ tk;sxh (D) dsUnzh; pedhyh fÝat dkyh fÝat gks tk;sxhA
9.
A Young’s double slit experiment is performed with white light, (A*) The central fringe will be white (B*) There will not be a completely dark fringe (C) The fringe next to the central will be red (D*) The fringe next to the central will be violet ;ax ds f}js[kkfNnz iz;ksx dks 'osr izdk'k }kjk fd;k tk;s rks (A*) dsUnzh; fÝat 'osr gksxhA (B*) iw.kZ dkyh fÝatas ugha gksxhA (C) dsUnzh; fÝat ds vxyh okyh yky gksxh (D*) dsUnzh; fÝat ds vxyh okyh cSx a uh gksxh
10.
Four light waves are represented by pkj izdk'k rjaxksa dks fuEu }kjk iznf'kZr (i) y = a1sin t. (ii) y = a2sin(t + ) (iii) y = a1sin 2t Interference frings may be observed due to superposition of
fd;k tkrk gS -
(iv) y = a2sin 2(t + )
O;frdj.k fÝats fuEu ds v/;kjksi.k ds dkj.k fn[kkbZ nsxh (A*) (i) and (ii)
(B) (i) and (iii)
(C) (ii) and (iv)
(D*) (iii) and (iv)
WORKED OUT EXAMPLES 1. Sol.
White light is a mixture of wavelengths between 400 nm and 700 nm. If this light goes through water ( = 1.33), what are the limits of the wavelength there? When a light having wavelength 0 in vacuum goes through a medium of refractive index , the wavelength in the medium becomes = 0/. For
0 = 400 nm, =
400 nm = 300 nm 1.33
700 nm = 525 nm. 1.33 Thus, the limits are 300 nm and 525 nm.
and for l0 = 700 nm, l =
2.
The optical path of a monochromatic light it same if it goes through 2.00 cm of glass or 2,25 cm of water. If the refractive index of water is 1.33, what is the refractive index of glass? manishkumarphysics.in
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Chapter # 17 Light Waves Sol. When light travels through a distance x in a medium of refractive index , its optical path is x. Thus, if is the refractive index of glass, (2.00 cm) = 1.33 × (2.25 cm) 2.25 = 1.50. 2.00 White light is passed through a double slit and interference pattern is observed on a screen 2.5 m away. The sepration between the slits is 0.5 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringe. Calculate the wavelengths of the violet and the red light. ,d f}&fNnz ls lQsn çdk'k dks fudkyk tkrk gS vkSj O;frdj.k&çfr:i insZ ij 2.5 ehVj dh nwjh ij ns[kk tkrk gSA fNnz (slits) ds e/; nwjh 0.5 mm gSA dsUnzh; lQsn fÝUt ls igyh cSaxuh rFkk yky fÝUt Øe'k% 2.0 mm rFkk 3.5 mm nwjh ij curh gSaA cSxa uh rFkk yky çdk'k ds rjaxnS/;Z dh x.kuk dhft, \ [HCV-1_Ch.-17_WOE_3]
or, 3.
Sol.
= 1.33 ×
For the first bright fringe, the distance from the centre is y =
D d
(2.5 m) (0.5 mm)(2.0mm) For violet light, y = 2.0 mm. Thus, 2.0 mm = 0.5 mm or = = 400 nm. 2.5 mm Similarly, for red light , y = 3.5 mm. Thus, 3.5 mm = 4.
(2.5m) or, = 700 nm. Ans. 0.5mm
A double slit experiment is performed with sodium (yellow) light of wavelength 589.3 nm and the interference pattern is observed on a screen 100 cm away. The tenth bright fringe has its centre at a distance of 12 mm from the central maximum. Find the separation between the slits. ,d f}&fNnz ç;ksx esa lksfM;e (ihyk) çdk'k ftldk rjaxnS/;Z 589.3 nm gS] dk mi;ksx fd;k tkrk gS rFkk 100 cm nwjh ij fLFkr insZ ij O;frdj.k çfr:i çkIr fd;k tkrk gSA nloha pedhyh fÝUt dk dsUnz] dsUnzh; mfPp"B ls 12 mm dh nwjh
ij gSA fNnzkas ds e/; nwjh crkg;sA Sol.
HCV_17_WOE_4
For the nth maximum fringe, the distance above the central line is x =
n D . d
According to the data given, x = 12 mm, n = 10, = 589.3 nm, D = 100 cm. Thus, the separation between the slits is 10 589.3 10 19 m 100 10 2 m n D = x 12 10 3 m = 4.9 × 10–4 m = 0.49 mm.
d=
5.
Sol.
The intensity of the light coming from one of the slits in a Young’s double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. The intensity of the light originating from the first slit is double the intensity from the second slit. The amplitudes of the two interfering waves are in the ratio
2 : 1 , say
2 A and A.
At the point of constructive interference, the resultant amplitude becomes ( 2 + 1)A. At the points of destructive interference, this amplitude is ( 2 – 1)A. The ratio of the resultant intensities at the maxima to that at the minima is
( 2 1)2 A 2 ( 2 1)2 A 2 6.
Sol.
3A
The width of one of the two slits in a Young’s double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit-width, find the ratio of the maximum to the minimum intensity in the interference pattern. Suppose the amplitude of the light wave coming from the narrower slit is A and that coming from the wider slit is 2A. The maximum intensity takes place. Then the resultant amplitude is the sum of the individual amplitudes. Thus, Amax = 2A + A = 3A The minimum intensity occurs at a place where destructive interference takes place. The resultant amplitude is then difference of the individual amplitudes. Thus, manishkumarphysics.in
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Chapter # 17
Light Waves Amin = 2A – A = A. As the intensity is proportional to the square of the amplitude,
max ( A max )2 (3 A )2 9. min ( A min )2 A2 7.
Two sources S1 and S2 emitting light of wavelength 600 nm are placed a distance 1.0 × 10–2 cm apart. A detector can be moved an the line S1P which is perpendicular to S1S1. (a) What would be the minimum and maximum path difference at the detector as it is moved along the line S1P? (b) Locate the position of the farthest minimum detected. S2 d
Sol. S1
P
D
(a) The situation is shown in figure. The path difference is maximum when the detector is just at the position of S1 and its value is equal to d = 1.0 × 10–2 cm. The path difference is minimum when the detector is at a large distance from S1. The path difference is then close to zero. (b) The farthest minimum occurs at a point P where the path difference is /2. If S1 P = D, S2 P – S1P =
2
D 2 d2 – D = 2
or,
or,
D + d = D 2
or,
d2 = D =
or,
D=
2
=
2
2
2 4
d2 – 4
(1.0 10 4 m)3 600 10 9 m
– 150 × 10–9 m = 1.7 cm.
8.
A beam of light consisting of two wavelengths, 6500 Å and 5200 Å is used to obtain slit experiment (1 Å = 10– 10 m). The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Sol.
(a) The centre of the nth bright fringe is at a distance y =
nD from the central maximum. For the 3rd bright d
fringe of 6500 Å, y=
3 6500 10 10 m 1.2 m
2 10 3 m = 0.117 cm = 0.12 cm. (b) Suppose the mth bright fringe of 6500 Å coincides with the nth bright fringe of 5200 Å.
Then,
m 6500 Å D n 5200 Å D = d d
5200 4 m = = . 6500 5 n The minimum values of m and n that satisfy this equation are 4 and 5 respectively. The distance of the 4th bright fringe of 6500 Å or the 5th bright fringe of 5200 Å from the central maximum is
or,
manishkumarphysics.in
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Chapter # 17
Light Waves
4 6500 10 10 m 1.2 m y=
2 10 3 m
= 0.156 cm 0.16 cm. 9.
Sol.
Monochromatic light of wavelength 600 nm is used in a Young’s double slit experiment. One of the slits is covered by a transparent sheet of thickness 1.8 × 10–5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet? When the light travels through a sheet of thickness t, the optical path travelled is t where is the refractive index. When one of the slits is covered by the sheet, air is replaced by the sheet and hence, the optical path changes by ( – 1)t. One fringe shifts when the optical path changes by one wavelength. Thus, the number of fringes shifts when the optical path changes by one wavelength. Thus, the number of fringes shifted due to the introduction of the sheet is ( 1)t (1.6 1) 1.8 10 5 m 18 = 600 10 9 m
10. Sol.
White light is incident normally on a glass plate of thickness 0.50 × 10–5 and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate? The light of wavelength l is strongly reflected if 1 2d = n , 2 where n is a nonnegative integer. Here, 2d = 2 × 1.50 × 0.5 × 10–6 m =1.5 × 10–6 m Putting l = 400 nm in (i) and using (ii),
.....(i)
....(ii)
1 1.5 × 10–6 m = n (400 × 10–9 m) 2 or, n = 3.25. Putting l = 700 nm is (ii), 1 1.5 × 10–6 m = n (700 × 10–9 m) 2 or, n = 1.66 Thus, within 400 nm to 700 nm the integer n can take the values 2 and 3. Putting these values of n in (i), the wavelengths become
4 d = 600 nm and 429 nm. 2n 1 Thus, light of wavelengths 429 nm and 600 nm are strongly reflected. =
11. Sol.
A parallel beam of green light of wavelength 546 nm passes through a slit of width 0.40 mm. The transmitted light is collected on a screen 40 cm away. Find the distance between the two first order minima. The minima occur at an angular deviation given by b sin = n, where n is an integer. For the first order minima, n = ± 1 so that sin = ±
D . b
Thus, the minima are formed at a distance
D from the central maximum on its two sides. The separation b
between the minima is 2 546 10 9 m 40 10 2 m 2D = = 1.1 mm. b 0.40 10 3 m
manishkumarphysics.in
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Chapter # 17
Light Waves
EXERCISE
1.
Find the range of frequency of light that is visible to an average human being (400 nm < < 700 nm). lk/kkj.k euq"; }kjk ns[kh tk ldus okyh izdk'k rjaxksa dh vko`fÙk Kkr dhft;sA (400 nm < < 700 nm) Ans. 774.3 × 1014 Hz- 7.5 × 1014 Hz
2.
The wavelength of sodium light in air is 589 nm. (a) Find its frequency in air. (b) Find its wavelength in water (refractive index - 1.33). (c) Find its frequency in water. (d) Find its speed in water. ok;q esa lksfM;e izdk'k dh rjaxnS/;Z 589 nm gS % (a) ok;q esa vko`fÙk Kkr dhft;sA (b) ty esa rjaxnS/;Z Kkr dhft;sA (viorZukad = 1.33) (c) ty esa vko`fÙk Kkr dhft;sA (d) ty esa bldh pky Kkr dhft;sA Ans. (a) 5.09 × 1014 hz (b) 443 nm (c) 5.09 × 1014 Hz (d) 2.25 × 108 m/s.
3.
The index of refraction of fused quartz is 1.472 for light of wavelength 400 nm and is 1.452 for light of wavelength 760 nm. Find the speeds of light of these wavelengths in fused quartz. 400 nm rjaxnS/;Z okys izdk'k ds fy;s ¶;wTM DokV~t Z dk viorZukad 1.472 gS rFkk 760 nm rjaxnS/;Z okys izdk'k ds fy;s 1.452 gSA bu rjaxnS/;ks± ds fy;s ¶;wTM DokV~Zt esa izdk'k dk osx Kkr dhft;sA Ans. 2.04 × 108 m/s, 2.07 × 108 m/s
4.
The speed of the yellow light in a certain liquid is 2.4 × 108 m/s. Find the refractive index of the liquid. k The speed of the yellow light in a certain liquid is 2.4 × 108 m/s. Find the refractive index of the liquid. k Ans. 1.25
5.
Two narrow slits emitting light in phase are separated by a distance of 1.0 cm. The wavelength of the light is 5.0 × 10–7 m. The interference pattern is observed on a screen placed at a distance of 1.0 m. (a) Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima? (b) Find the separation between the sources which will give a separation of 1.0 mm between the consecutive maxima. nks lehi okys fNnzkas (ftuds chp dh nwjh 1.0 cm gS) esa ls çdk'k fudyrk gSA çdk'k dh rjaxnS/;Z 5.0 × 10–7 m gSA 1.0 m dh nwjh ij j[ks insZ ij O;frdj.k çfr:i çkIr fd;k tkrk gSA (a) nks Øekxr mfPp"Bksa ds chp dh nwjh Kkr dhft,A D;k vki ;g vk'kk dj ldrs gSa fd bu mfPp"Bksa dks vyx&vyx ns[k ldsAa (b) mu lzksrksa ds e/; dh nwjh Kkr dhft, tks nzO;eku vf/kdre ds chp 1.0 mm dh nwjh nssrs gSA HCV_Ch-17_Ex._5 Ans. (a) 0.05 mm (b) 0.50 mm
6.
The separation between the consecutive dark fringes in a Young’s double slits experiment is 1.0 mm. The screen is placed at a distance of 2.5 m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment. ;ax f}fNnz ç;ksx esa nks Øekxr vnhIr fÝUtksa ds chp dh nwjh 1.0 mm gSA fNnzkas ls insZ ds chp dh nwjh 2.5 m gS rFkk fNnzkas ds chp dh nwjh 1.0 mm gSA ç;ksx esa ç;qDr çdk'k dk rjaxnS/;Z crkb;sA HCV_Ch-17_Ex_6 Ans.400 nm
7.
In a double slit inteference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10–7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centrimeter width on the screen ? f}fNnz ç;ksx esa fNnzkas ds e/; dh nwjh 1.0 mm, gS rFkk çdk'k dk rjaxnS/;Z 5.0 × 10–7 ehVj gSA fNnzkas ls insZ ds chp dh nwjh 1.0 ehVj gSA (a) çFke fufEu"B ds dsUnz ls dsUnzh; mfPp"B ds dsUnz ds chp dh nwjh dh x.kuk dhft,A (b) insZ dh 1 lseh0 pkSM+kbZ esa fdruh pedhyh fÝUt cusxh ? HCV_Ch-17_Ex_7 Ans. (a) 0.25mm (b) 20
8.
In a Young’s double slits experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away ? ;ax f}&fLyV iz;ksx esa] nks m/okZ/kj ladjh fLyVsa 0.800 nm nwjh ij j[kh xbZ gS] budks 589 nm rjaxnS/;Z okys ihys izdk'k ds ,d gh lzkrs ls vkyksfdr fd;k x;k gSA 2.00 eh- nwj fLFkr insZ ij izfs {kr O;frdj.k izfr:i esa nks Øekxr pedhyh fÝatksa
ds e/; fdruh nwjh gS\ Ans. 1.47 mm
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Chapter # 17 Light Waves 9. Find the angular separation between the consecutive bright fringes in a Young’s double slits experiment with blue-green light of wavelength 500 nm. The separation between the slits is 2.0 × 10–3 m. ;ax f}&fLyV iz;ksx esa] uhyk&gjk izdk'k iz;qDr fd;k x;k gSA ftldh rjaxnS/;Z 500 uSuksehVj gSA js[kkfNnzksa ds e/; 2.0 × 10–3 eh- varjky gSA nks Øekxr pedhyh fÝatksa ds e/; dks.kh; varjky Kkr dhft;sA Ans. 0.014 degree 10.
A source emitting light of wavelengths 480 nm and 600 nm is used in a double slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum correspoinding to the two wavelengths. f}&fLyV iz;ksx esa iz;D q r izdk'k lzkrs 480 nm rFkk 600 nm rjaxnS/;ks± dk izdk'k mRlftZr dj jgk gSA fLyVksa ds e/; varjky 0.25 feeh gS rFkk O;frdj.k izfr:i dk iz{s k.k fLyVksa ls 150 lseh nwj fLFkr insZ ij fd;k tkrk gSA nksuksa rjaxnS/;ks± ds laxr
izFke] mfPp"Bksa dk jSf[kd varjky ¼dsUnzh; mfPp"Bksa ls vxys mfPp"Bksa dk½ Kkr dhft;sA Ans. 0.72 mm 11.
White light is used in a Young’s double slit experiment. Find the minimum order of the violet fringe ( = 400 nm) which overlaps with a red fringe ( = 700 nm). ;ax&f}fNnz ç;ksx esa lQsn çdk'k dk mi;ksx fd;k tkrk gSA cSaxuh fÝUt ( = 400 nm) ds fy, og U;wure Øe fudkfy, tc ;g yky fÝUt ij ( = 700 nm) lEikrh gks tk,A HCV_Ch-17_Ex_11 Ans. 7
12.
Find the thickness of a plate which will produce a change in optical path equal to half the wavelength of the light passing through it normally. The refractive index of the plate is µ. fdlh ifV~Vdk dh og eksVkbZ Kkr dhft;s] tks bl ij vfHkyEcor~ vkifrr izdk'k fdj.k esa ckgj fudyus ij rjaxnS/;Z ds vk/ks eku dk izdkf'kd iFk ifjorZu mRiUu dj nsAa ifV~Vdk dk viorZukad µ gSA Ans.
13.
2(u 1)
A plate of thickness t made of a material of refractive index µ is placed in front of one of the slits in a dboule slits experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero. Wavelength of the light used is . Neglect any absorption of light in the plate. ;ax f}&fLyV iz;ksx esa ,d fLyV ds lkeus µ viorZukad okys inkFkZ dh t eksVkbZ dh ifV~Vdk j[k nh tkrh gSA (a) ifV~Vdk j[kus ds dkj.k izdkf'kd iFk esa ifjorZu Kkr dhft;sA (b) ifV~Vdk dh U;wure eksVkbZ t fdruh gksuh pkfg;s fd fÝat izfr:i ds dsUnz ij rhozrk 'kwU; gks\ iz;qDr izdk'k dh rjaxnS/;Z gSA ifV~Vdk }kjk izdk'k dk vo'kks"k.k ux.; gSA Ans. (a) (µ – 1) t,
14.
(b)
2(u 1)
A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed? ;ax f}&fLyV iz;ksx esa 620 nm rjaxnS/;Z dk ,d o.khZ izdk'k iz;D q r gks jgk gS] bldh ,d fLyV ds lkeus 0.02 mm eksVkbZ dk ,d ikjn'kZd dkxt (viorZukad = 1.45) fpidk;k x;k gSA ;fn ;g dkxt gVk fy;k tk;s rks dsUnz ls fdruh fÝats
ikj gksxh\
Ans. 14.5 15.
In a Young’s double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron ( 1 micron = 10–6 m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maximum now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment. ;ax f}&fLyV iz;ksx esa ,d o.khZ izdk'k dk iz;ksx fd;k tk jgk gS] tc 1.6 viorZukad okyh vHkzd dh 1.964 ekbØksu ( 1 ekbØksu = 10–6 eh-) eksVkbZ dh ifV~Vdk fdl ,d O;frdkfjr rjax ds ekx esa j[k nh tkrh gS] rks insZ ij fÝat izfr:i dqN
fuf'pr nwjh ls foLFkkfir gks tkrk gSA vc vHkzd dh ifV~Vdk gVk nh tkrh gS ,oa insZ rFkk fLyVksa ds chp dh nwjh nqxuh dj nh tkrh gSA ;g izfs {kr fd;k tkr gS fd vc nks Øekxr mfPp"Bksa ds chp dh nwjh mruh gh gS] ftruh vHkzd&ifV~Vdk izfo"V djkus ij izkIr gqbZ FkhA iz;ksx esa iz;qDr ,do.khZ izdk'k dh rjaxnS/;Z dh x.kuk dhft;sA Ans. 590 nm manishkumarphysics.in
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Chapter # 17 Light Waves 16. A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
,d vHkzd dh ifV~Vdk rFkk ,d ikWfyLVj dh ifV~Vdk f}&fLyV iz;ksx midj.k dh nksuksa fLyVksa ij yxk nh x;h gSA ifV~Vdkvksa dh eksVkbZ 0.50 mm rFkk fLyVksa dk vUrjky 0.12 cm gSA 590 nm rjaxnS/;Z ds izdk'k ds fy,] tks fd iz;ksx esa iz;D q r fd;k x;kA vHkzd rFkk ikWfyLVj ds viorZukad Øe'k% 1.58 rFkk 1.55 gSA O;frdj.k izfr:i dk iz{s k.k 1 eh- nwj fLFkr ins± ij fd;k tk jgk gSA (a) fÝat varjky fdruk gS\ (b) dsUnz ls fdruh nwjh ij izFke mfPp"B fLFkr gksxk\ Ans. (a) 4.9 × 10–4 m 17.
(b) 0.021 cm on one side and 0.028 cm on the other side
Two transparent slabs having equal thickness but different refractive indices µ1 and µ2 are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young’s experiment so that the light from one slit goes through one material and the light from the other slit goes through one material and the light from the other slit goes through the other material. What should be the minimum non zero thickness of the slab so that there is maximum at the point P0 which is equidistant from the slits ? leku eksVkbZ fdUrq fHkUu&fHkUu viorZukad µ1 ,oa µ2 okyh nks ikjn'kZd ifV~Vdk,¡ ik'oZr% fpidk dj ,d la;qDr ifV~Vdk
cuk;h tkrh gSA ;g la;D q r ifV~Vdk dk ;ax f}&fLyV iz;ksx esa fLyVksa ds lkeus bl izdkj j[kh tkrh gS fd ,d fLyVksa ds lkeus bl izdkj j[kh tkrh gS fd ,d fLyV dk izdk'k ,d inkFkZ ls gksdj xqtjrk gSA ifV~Vdk dh U;wure eksVkbZ fdruh gksuh pkfg;s] ftlls fd fLyVksa ls leku nwjh ij fLFkr fcUnq P0 ij ,d fufEu"B izkIr gks\ Ans. 2 | µ µ | 1 2 18.
A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young’s double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pased on the other slit also? The wavelength of the light used is 600 nm. 0.02 mm eksVkbZ ds ,d irys dkxt dk viorZukad 1.45 gS] bldks ;ax f}&fLyV iz;ksx esa ,d fLyV ds lkeus fpidk fn;k x;k gSA dkxt bl ij vkifrr izdk'k ÅtkZ dk 4/9 Hkkx ikjxfer dj nsrk gSA (a) fÝat izfr:i esa vf/kdre rhozrk rFkk U;wure rhozrk dk vuqikr Kkr dhft;sA (b) ;fn ,d ,slk gh dkxt fLyV ij Hkh fpidk fn;k tk;s rks dsUnz ls fdruh fÝats ikj gksxh\ iz;qDr izdk'k dh rjaxnS/;Z 600 nm gSA Ans. (a) 25 (b) 15
19.
A Young’s double slit apparatus has slits separated by 0.28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light ( = 700 nm is vacuum). Find the fringe-width of the pattern formed on the screen. ;ax f}&fLyV iz;ksx esa fLyVksa dk varjky 0.28 mm gS rFkk fLyVksa ls insZ dh nwjh 48 lseh gSA lEiw.kZ midj.k ikuh esa Mqcks fn;k tkrk gS rFkk fLyVksa dks yky izdk'k ( = 700 nm fuokZr es)a ls izdkf'kr fd;k tkrk gSA insZ ij cuus okys fÝat izfr:i
dk fÝat&varjky Kkr dhft;sA Ans. 0.90 mm 20.
A parallel beam of monochromatic light is used in a Young’s double slit experiment. The siits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle = sin–1 with the normal to the plane of the slits, there will be a dark fringe at the 2d
centre P0 of the pattern.
;ax f}&fLyV iz;ksx esa ,d o.khZ izdk'k dk lekUrj iqt a iz;D q r fd;k tk jgk gSA insZ dh fLyVksa ls nwjh d gS rFkk fLyVksa ds 2d
lekukarj j[kk gqvk gSA O;Dr dhft;s fd ;fn vkifrr iqat] insZ ds ry ij vfHkyEc ls = sin–1
dks.k cukrk gS]
rks fÝat izfr:i ds dsUnz P0 ij dkyh fÝat izkIr gksxhA 21.
A narrow slit S transmitting light of wavelength is placed a distance d above a large plane mirror as shown in figure. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance D from the slit. (a) What will be intensity at a point just above the mirro, i.e., just above O? (b) At what distance from O does the first maximum occur? manishkumarphysics.in
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Chapter # 17
Light Waves
,d ladjh fLyV S ls rjaxnS/;Z dk izdk'k ikjxfer gks jgk gSA bldks fp=kkuqlkj ,d cgqr cM+s lery niZ.k ls d nwjh Åij j[kk x;k gSA fLyV ls vk jgk lh/kk izdk'k rFkk niZ.k ls ijkofrZr gksdj vkus okyk izdk'k] fLyV ls D nwjh ij fLFkr insZ ij O;frdj.k mRiUu djrs gSaA (a) niZ.k ls rqjar Åij fLFkr fcUnq ij vFkkZr~ rqjar Åij fcUnq O ij] rhozrk fdruh gksxh\ (b) O ls fdruh nwjh ij izFke mfPp"B izkIr gksxk\
Ans. (a) zero 22.
(b)
D 4d
A long narrow horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen 1.0 m away from the slit. Find the fringe-width if the light used has a wavelength of 700 nm. ,d {kSfrt lery niZ.k ls 1 feeh Åij ,d yEch {ksfrt fLyV j[kh x;h gSA fLyV ls 1.0 m nwj fLFkr insZ ij] fLyV ls lh/ks vkus okys rFkk ijkorZu ds i'pkr~ vkus okys izdk'k esa O;kfrdj.k gksrk gSA ;fn iz;D q r izdk'k dh rjaxnS/;Z 700 nm
gS] fÝat pkSM+kbZ Kkr dhft;sA Ans. 0.35 mm 23.
Consider the situation of the previous problem. If the mirror reflects only 64% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen? fiNys iz'u esa of.kZr fLFkfr ij fopkj dhft;sA ;fn niZ.k bl ij vkifrr izdk'k ÅtkZ dk 64% gh ijkofrZr djrk gS] insZ
ij izfs {kr O;frdj.k izfr:i esa vf/kdre rhozrk rFkk U;wure rhozrk dk vuqikr fdruk gksxk\ Ans. 81 : 1 24.
A double slit S1 – S2 in illuminated by a coherent light of wavelength . The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D1 from it and a screen is placed behind the double slit at a distance D2 from it (figure). The screen receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen. ,d f}&fLyV S1 – S2 , rjaxnS/;Z okys dyk lEc) izdk'k ls izdkf'kr gSA fLyVksa ds e/; nwjh d gSA fLyVksa ds lkeus D1 nwjh ij ,d lery niZ.k j[kk gqvk gS rFkk f}&fLyV ds ihNs dh vksj buls D2 nwjh ij ,d inkZ j[kk gqvk gS ¼fp=k½A inkZ
dsoy niZ.k ls ijkofrZr izdk'k gh xzg.k djrk gSA insZ ij izkIr foorZu izfr:i dk fÝat&varjky Kkr dhft;sA
Ans. (2D1 + D2)/d 25.
White coherent light (400 nm-700) is sent through the slits of a Young’s double slit experiment (as shown in the figure). The separation betweem the slits is 0.5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole ? (b) which wavelength(s) will have a strong intensity? ;ax f}&fLyV iz;ksx dh fLyVksa ls 'osr izdk'k fuxZfer gksrk gSA fLyVksa ds e/; varjky 0.5 feeh gS rFkk inkZ fLyVksa ls 50 lseh nwj gSA insZ ij dsUnzh; js[kk ls 1.0 mm nwj ¼fÝatksa dh pkSM+kbZ ds vuqfn'k½ ,d fNnz gSA (a) fNnz ls vkus okys izdk'k esa dkSulh rjaxnS/;Z vuqifLFkr gS\ (b) dkSulh rjaxnS/;Z ¼rjaxnS/;ks±½ dh rhozrk izcyre gS\
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Chapter # 17 Ans. (a) 400 nm, 667 nm, (b) 500 nm 26.
Consider the arrangement show in figure. The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe-width. fp=k esa iznf'kZr O;oLFkk ij fopkj dhft;sA fLyVksa ds e/; varjky d dk rqyuk esa nwjh D cgqr vf/kd gSA (a) d dk og U;wure eku Kkr dhft;s ftlls O ij dkyh fÝat izkIr gksA (b) ekuk fd d dk ;g eku gSA og nwjh x Kkr dhft;s] tgk¡ vxyh pedhyh fÝat cusxhA (c) fÝat varjky Kkr dhft;sA
Ans. (a) 27.
Light Waves
D 2
(b) d
(c) 2d
Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength . The separation between the sources is 2. Consider a line passing through S2 and perpendicular to the line S1 S2. What is the smallest distance from S2 where a minimum of intensity occurs? leku dyk esa dEiUu dj jgs nks fcUnq lzksr S1 rFkk S2 rjaxnS/;Z dk izdk'k mRlftZr dj jgs gSAa lzkrs ksa ds e/; nwjh 2 gSA S2 ls xqtjus okyh rFkk S1 S2 ds yEcor~ ,d js[kk ij fopkj dhft;sA S2 ls og U;wure nwjh fdruh gS] tgk¡ ij U;wure
rhozrk izkIr gksrh gSA Ans. 7/12 28.
As shown in the figure three equidistant slits being illuminated by a monochromatic parallel bea, of light. Let BP0 – AP0 = /3 and D >> . (a) Show that in this case d =
2D / 3 . (b) Show thast the intensity at P0 is
three times the intensity due to any of the three slits individually.
fp=k esa ,d o.khZ izdk'k ds lekukarj iqat ls izdkf'kr leku nwjh ij fLFkr rhu fLyVsa n'kkZ;h x;h gSA ekuk fd BP0 – AP0 = /3 rFkk D >> gSA (a) O;Dr dhft;s fd bl fLFkfr esa d = 2D / 3 gSA (b) O;Dr dhft;s fd P0 ij rhozrk] fdlh ,d vdsyh fLyV ds dkj.k rhozrk dh rhu xquh gSA
29.
In a Young’s double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is 0.20 W/m2, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes ? ;ax f}&fLyV iz;ksx esa] fLyVksa dk varjky = 2.0 feeh] izdk'k dh rjaxnS/;Z = 600 nm rFkk fLyVksa ls insZ dh nwjh = 2.0 eh- gSA ;fn dsUnzh; mfPp"B ds dsUnz ij rhozrk 0.20 okWV@eh2, gS] bl dsUnz ls] fÝat&pkSM+kbZ ds vuqfn'k 0.5 lseh nwjh ij
fLFkr fcUnq ij rhozrk fdruh gksxh\ Ans. 0.05 W/m2 30.
If a Young’s double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits . The slits are separated by the distance d and are illuminated by monochromatic light of wavelength . Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum. ;ax ds f}&fLyV O;frdj.k iz;ksx esa fLyVksa ls D nwjh ij fLFkr insZ ij] fÝat izfr:i izfs {kr fd;k tkrk gSA fLyVksa ds chp dh nwjh d gS rFkk budks rjaxnS/;Z okys ,do.khZ izdk'k ls izdkf'kr fd;k x;k gSA dsUnzh; fcUnq ls og nwjh Kkr dhft;s] tgk¡ rhozrk de gksdj] gks tkrh gS % (a) vf/kdre dh vk/kh, (b) vf/kdre dh ,d pkSFkkbZ Ans. (a)
D 2d
(b)
D 3d
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Chapter # 17 31.
Light Waves
In a Young’s double slit experiment = 500 nm, d = 1.0 mm and D = 1.0m. Find the minimum distance from the central maximum from the central maximum for which the intensity is half of the maximum intensity. ;ax f}&fLyV iz;ksx esa = 500 nm , d = 1.0 mm rFkk D = 1.0 m. gSA dsUnzh; mfPp"B ls og U;wure nwjh Kkr dhft;s]
ftlds fy;s rhozrk] vf/kdre rhozrk dh vk/kh gSA Ans. 1.25 × 10–4 m 32.
The linewidth of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the linewidth of a bright finge in a Young’s double slit experiment in terms of , d and D where the symbols have their usual meanings.
dHkh&dHkh pedhyh fÝat dh jSf[kd pkSMk+ bZ bl izdkj ifjHkkf"kr dh tkrh gS fd ^^fÝat dh pkSMk+ bZ dsUnzh; js[kk ds nksuksa vksj mu fcUnqvksa ds chp dh nwjh gS] tgk¡ rhozrk vf/kdre dh vk/kh jg tkrh gSA** ;ax ds f}&fLyV iz;ksx esa , d rFkk D ds inksa esa pedhyh fÝat dh jSf[kd pkSM+kbZ Kkr dhft;sA tgk¡ ladrs ksa dk lkekU; vFkZ gSA Ans. 33.
D 2d
Consider the situation shown in figure. The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength . The separation between the slits is d. The light transmitted by the slits falls on a screen 1 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3. another screen 2 is placed a further distance D away from 1. Find the ratio of the maximum to minimum intensity observed on 2 if z is equal to fp=k esa iznf'kZr fLFkfr ij fopkj dhft;sA dsUnzh; js[kk ls lefer j[kh gqbZ nks fLyVsa S1 rFkk S2 , rjaxnS/;Z okys ,do.khZ izdk'k ls izdkf'kr gSA fLyVksa ds e/; nwjh d gSA fLyVksa ls fuxZfer izdk'k] buls D nwjh ij j[ks gq, insZ 1 ij vkifrr gksrk gSA fLyV S3 dsUnzh; js[kk ij gS rFkk S3 ls z nwjh ij fLyV S4 fLFkr gSA 1 ls D nwjh ij ,d vU; inkZ 2 j[kk x;k gSA 2 ij izkIr vf/kdre rFkk U;wure rhozrkvksa dk vuqikr Kkr dhft;sA ;fn z rqY; gS : (a)
D d
Ans. (a) 1 34.
(b)
D 4d
(b)
(c) z =
D 2d
(c) 34
Consider the arrangement shown in figure. By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P which is at the common perpendicular bisector of S1S2 and S3S4. When z =
D , the intensity measured at P is . Find this intensity when z is equal to 2d
fp=k esa iznf'kZr O;oLFkk ij fopkj dhft;sA fdlh ;kaf=kd O;oLFkk ls fLyVksa S3 o S4 ds e/; dk varjky ifjofrZr fd;k tk ldrk gSA rhozrk dk ekiu P ij fd;k tkrk gS] tks S1S2 rFkk S3S4 tc z =
D , gS] P ij 2d
ekih x;h rhozrk gSA bl rhozrk
dk eku dhft;s] tc z dk eku gS :
D d Ans. (a) zero
(a)
35.
3D 2d (C) 2
(b) (b)
and
(c)
2D d
A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm. What is the index of refraction of the soap solution, if it is known to be between 1.2 and 1.5? 0.0011 feeh eksVkbZ okyh lkcqu dh fQYe 580 uSuksehVj rjaxnS/;Z okys ijkofrZr izdk'k esa dkyh fn[kkbZ nsrh gSA lkcqu ds manishkumarphysics.in
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Chapter # 17
Light Waves
?kksy dk viorZukad fdruk gS] ;fn ;g Kkr gks fd bldk eku 1.2 rFkk 1.5 ds chp esa gS\ Ans. 1.32 36.
A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index = 1.4). What should be the minimum thickness of the film so that it strongly reflects the light? 560 nm rjaxnS/;Z okyk ,d izdk'k iqt a ] rsy dh ,d iryh fQYe ¼viorZukad = 1.4½ ij vkifrr gSA fQYe dh U;wure
eksVkbZ fdruh gks fd ;g izdk'k dks izcyre :i ls ijkofrZr djsa\ Ans. 100 nm 37.
A parallel beam of white light is incident normally on a water film 1.0 × 10–4 cm thick. Find the wavelength in the visible range (400 nm-700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33. ikuh dh 1.0 × 10–4 lseh eksVh fQYe ij 'osr izdk'k dk lekukarj iqt a vfHkyEcor~ vkifrr gSA n`'; izdk'k ijkl (400 nm700 nm) esa fQYe }kjk rh{.ke viofrZr rjaxnS/;Z Kkr dhft;sA ikuh dk viorZukad = 1.33 Ans. 443 nm, 532 nm and 666 nm
38.
A glass surface is coated by an oil film of uniform thickness 1.00 × 10–4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. find the wavelengths of light in the visible region (400nm-750nm) which are weakly transmitted by the oil film under normal incidence. dkap dh lrg ij ,d leku eksVkbZ 1.00 × 10–4 lseh dh rsy ijr yxk nh xbZ gSA rsy dk viorZukad 1.25 rFkk dkap dk 1.50 gSA n`'; izdk'k {ks=k (400 nm - 750nm) esa izdk'k dh mu rjaxnS/;ks± ds eku Kkr dhft;s] tks vfHkyEcor~ vkiru
dh fLFkfr esa rsy&fQYe ls iw.kZr;k ikjxfer gks tk;sxhA Ans. 455 nm, 556 nm, 714 nm 39.
Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at = 30°. 5.0 lseh pkSM+kbZ dh ,d yEch fLyV ij lery lw{e&rjaxas vkifrr gSA ;fn izFke foorZu fufEu"B = 30° ij izkIr gksrk
gS] rks lw{e&rjaxksa dh rjaxnS/;Z dh x.kuk dhft;sA Ans. 2.5 cm 40.
Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and falls on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall? 560 nm rjaxnS/;Z dk izdk'k 0.20 mm f=kT;k ds lwph&fNnz (pinhole) ls xqtjdj 2.00 eh nwj fLFkr nhokj ij vkifrr
gksrk gSA nhokj ij cuus okys dsUnzh; pedhys /kCcs dh f=kT;k fdruh gksxh\ Ans. 1.37 cm 41.
A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed? 620 nm rjaxnS/;Z okys izdk'k ds lekukarj iqt a dks Qksdflr djus ds fy;s 8.0 lseh O;kl okys mÙky ySl a dk mi;ksx fd;k tkrk gS] ;fn izdk'k ySl a ls 20 lseh nwj Qksdflr gksrk gS] cuus okys dsUnzh; pedhys /kCcs dh f=kT;k fdruh gksxh\ Ans. 3.8 × 10–6 m
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