Wave Optics
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Four identical monochromatic sources A, B, C, D as shown in the (figure)
produce waves of the same wavelength λ and are coherent. Two receiver
R1 and R2 are at great but equal distances from B.
(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when B is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?
(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?
Four identical monochromatic sources A, B, C, D as shown in the (figure)
produce waves of the same wavelength λ and are coherent. Two receiver
R1 and R2 are at great but equal distances from B.
(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when B is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?
(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation-consider the disturbance at the receiver R1 which is at a distance d from B
YA= acos(wt) and path difference is hence phase difference is .
Thus the wave R1 because of B
YB= acos(wt- )= - acoswt here path d
Upvote 
A small transparent slab containing material of µ = 1.5 is placed along AS2(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?
A small transparent slab containing material of µ = 1.5 is placed along AS2(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- as the refractive index of the class , the path difference will be calculated as =2dsin +( )L
For principal maxima ,(path difference is zero)
2dsin 0+( )L=0
Sin 0= - =
Sin 0=-1/16
OP=Dtan 0= Dsin 0=-D/16
For p
Upvote Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I 0 𝐼 0 is the intensity of the principal maxima when no polariser is present, calculte in the present case, the intensity of the principal maxima as well as the first minima.
Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I 0 𝐼 0 is the intensity of the principal maxima when no polariser is present, calculte in the present case, the intensity of the principal maxima as well as the first minima.
This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- when polariser is not used
A=Aperp+A
letA1= asinwt and A2=asin(wt+ )
now superposition principle for perpendicular polariser
AR= asinwt+ asin(wt+ )
AR=a(2cos sin(wt+ ))
AR=2acos sin(wt+ )
This eqn is also same for
Upvote 10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n /a.
Justify this by suitably dividing the slit to bring out the cancellation.
10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n /a.
Justify this by suitably dividing the slit to bring out the cancellation.
10.21 Consider that a single slit of width is divided in to smaller slits.
Therefore width of each slit,
Angle of diffraction is given by the relation,
=
Now, each of these infinitesimally small slit sends zero intensity in the direction of Ø.
Hence, the combination of these
Upvote 10.20 Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
10.20 Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
10.20 Weak radar signals sent by a low flying aircraft can interfere with the TV signal received by the antenna. Hence, TV signal may get distorted, resulting in shaking of picture on the TV.
This is because superposition follows from the linear character of a differential equation that governs wave
Upvote 10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
10.19 Wavelength of the light beam, = 500 nm = 500 m
Distance of the screen from the slit, D = 1 m
Distance of the first minimum from the centre of the screen, x = 2.5 mm = 2.5 m
Let the width of the slit be = d
From the equation
we get d =
= 2 m = 0.2 mm
Hence, the width of the slot is
Upvote 10.18 Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
10.18 Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
10.18 Distance between the towers, d = 40 km
Height of the line joining the hills, h = 50 m
Thus, the radial spread of the radio wave should not exceed 40 km
Since the hill is located halfway between the towers,
Fresnel's distance = = 20 km = 2
Aperture can be taken as a = h = 50 m
Fresnel's d
Upvote 10.17 Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
10.17 Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
10.17 If the width of the slit is made double the original width, then the size of the central diffraction band reduces to half and the intensity increases up to four times.
The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of t
Upvote 10.16 In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?
10.16 In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?
10.16 Wavelength of the light used, 6000 nm = 600 m
Angular width of the fringe, = 0.1 rad
Angular width of a fringe is related to slit spacing (d) as
d = = = 3.44 m
Upvote 10.15 For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
10.15 For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
10.15 Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.
In case of light waves, so
Upvote