Consider that a single slit of width d is divided into n smaller slits. Therefore, width of each slit, d' = d/n Angle of diffraction is given by the relation, 0 = (d λ/d') /d = λ/d' Now, each of these infinitesimally small slit sends zero intensity in direction 0. Hence, the combination of these sliRead more
Consider that a single slit of width d is divided into n smaller slits.
Therefore, width of each slit,
d’ = d/n
Angle of diffraction is given by the relation,
0 = (d λ/d’) /d = λ/d’
Now, each of these infinitesimally small slit sends zero intensity in direction 0. Hence, the combination of these slits will give zero intensity.
Ans (a). Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Ans (b). The prRead more
Ans (a).
Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen.
Ans (b).
The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second order wave equation, then any linear combination of y1 and y2 will also be the solution of the wave equation.
Wavelength of light beam, λ= 500 nm = 500 x 10⁻9 m Distance of the screen from the slit, D = 1 m For first minima, n = 1 Distance between the slits = d Distance of the first minimum from the centre of the screen can be obtained as: x = 2.5 mm = 2.5 x 10-3 m It is related to the order of minima as: dRead more
Wavelength of light beam, λ= 500 nm = 500 x 10⁻9 m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 x 10-3 m
It is related to the order of minima as: d
nλ = x d/D
d= nλ D/x
= 1 x 500 x 10⁻9 x 1/(2.5 x 10-3 ) = 2 x 10-4 m= 0.2 mm
Distance between the towers, d = 40 km Height of the line joining the hills, d = 50 m. Thus, the radial spread of the radio waves should not exceed 50 km. Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as: Zp = 20 km = 2 x 104 m Aperture can be taken as: a =Read more
Distance between the towers, d = 40 km
Height of the line joining the hills, d = 50 m.
Thus, the radial spread of the radio waves should not exceed 50 km.
Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:
Zp = 20 km = 2 x 104 m
Aperture can be taken as: a = d = 50 m
Fresnel’s distance is given by the relation, Zp =a²/λ
Where, λ = Wavelength of radio waves ,
Therefore , λ=a²/Zp
= (50)²/ 2 x 104
= 1250 x 10⁻⁴ = 0.1250 m = 12.5 cm
(50)’ .
= -^—4-= 1250×10 4 =0.1250 m = 12.5 cm 2×10
Therefore, the wavelength of the radio waves is 12.5 cm.
Ans (a). In a single slit diffraction experiment, 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 of the central diffraction band increases up to four times. Ans (b). The interference pattern in a double-slitRead more
Ans (a).
In a single slit diffraction experiment, 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 of the central diffraction band increases up to four times.
Ans (b).
The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
Ans (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. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
Ans (d).
Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.
On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
Ans (e).
The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.
Wavelength of light used, λ = 6000 nm = 600 x 10-9 m Angular width of fringe, 0 = 0.1° = 0.1 x π/180 = 3.14 /1800 rad Angular width of a fringe is related to slit spacing (d) as: 0 = λ /d => d = λ/0 => d = (600 x 10-9) /(3.14/1800) = 3.44 x 10⁻4 m Therefore, the spacing between the slits isRead more
Wavelength of light used, λ = 6000 nm = 600 x 10-9 m
Angular width of fringe, 0 = 0.1° = 0.1 x π/180 = 3.14 /1800 rad
Angular width of a fringe is related to slit spacing (d) as: 0 = λ /d => d = λ/0
=> d = (600 x 10-9) /(3.14/1800) = 3.44 x 10⁻4 m
Therefore, the spacing between the slits is 3.44 x 10⁻4 m.
No 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, sounRead more
No
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, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.
Ans (a). The speed of light in a vacuum i.e., 3 x 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum. Ans (b). Out of the listed factors, the speed of light inRead more
Ans (a).
The speed of light in a vacuum i.e., 3 x 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.
Ans (b).
Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.
No; Wave theory Newton's corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the comRead more
No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.
Hence, we can write the expression:
c sin i = v sin r … (i)
Where, i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as: μ =v/c
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v. The wave picture of light is consistent with the experimental results.
Wavelength of Ha line emitted by hydrogen, λ = 6563 Aº = 6563 x 10⁻10 m. Star’s red-shift, (λ ' —λ ) = 15 Aº = 15 x 10-10 m Speed of light, c = 3 x 108 m/s Let the velocity of the star receding away from the Earth be v. The red shift is related with velocity as: λ ' —λ = v λ /c v = c/λ x (λ ' —λRead more
Wavelength of Ha line emitted by hydrogen, λ = 6563 Aº = 6563 x 10⁻10 m.
Star’s red-shift, (λ ‘ —λ ) = 15 Aº = 15 x 10-10 m
Speed of light, c = 3 x 108 m/s
Let the velocity of the star receding away from the Earth be v.
The red shift is related with velocity as:
λ ‘ —λ = v λ /c
v = c/λ x (λ ‘ —λ )
= (3 x 108 x 15 x 10-10 )/(6563 x 10⁻10) =6.87 x 10⁵
Therefore, the speed with which the star is receding away from the Earth is 6.87 x 10⁵ m/s.
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.
Consider that a single slit of width d is divided into n smaller slits. Therefore, width of each slit, d' = d/n Angle of diffraction is given by the relation, 0 = (d λ/d') /d = λ/d' Now, each of these infinitesimally small slit sends zero intensity in direction 0. Hence, the combination of these sliRead more
Consider that a single slit of width d is divided into n smaller slits.
Therefore, width of each slit,
d’ = d/n
Angle of diffraction is given by the relation,
0 = (d λ/d’) /d = λ/d’
Now, each of these infinitesimally small slit sends zero intensity in direction 0. Hence, the combination of these slits will give zero intensity.
See lessAnswer 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?
Ans (a). Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Ans (b). The prRead more
Ans (a).
Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen.
Ans (b).
The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second order wave equation, then any linear combination of y1 and y2 will also be the solution of the wave equation.
See lessA 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.
Wavelength of light beam, λ= 500 nm = 500 x 10⁻9 m Distance of the screen from the slit, D = 1 m For first minima, n = 1 Distance between the slits = d Distance of the first minimum from the centre of the screen can be obtained as: x = 2.5 mm = 2.5 x 10-3 m It is related to the order of minima as: dRead more
Wavelength of light beam, λ= 500 nm = 500 x 10⁻9 m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 x 10-3 m
It is related to the order of minima as: d
nλ = x d/D
d= nλ D/x
= 1 x 500 x 10⁻9 x 1/(2.5 x 10-3 ) = 2 x 10-4 m= 0.2 mm
Therefore, the width of the slits is 0.2 mm
See lessTwo 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?
Distance between the towers, d = 40 km Height of the line joining the hills, d = 50 m. Thus, the radial spread of the radio waves should not exceed 50 km. Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as: Zp = 20 km = 2 x 104 m Aperture can be taken as: a =Read more
Distance between the towers, d = 40 km
Height of the line joining the hills, d = 50 m.
Thus, the radial spread of the radio waves should not exceed 50 km.
Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as:
Zp = 20 km = 2 x 104 m
Aperture can be taken as: a = d = 50 m
Fresnel’s distance is given by the relation,
Zp =a²/λ
Where, λ = Wavelength of radio waves ,
Therefore , λ =a²/Zp
= (50)²/ 2 x 104
= 1250 x 10⁻⁴ = 0.1250 m = 12.5 cm
(50)’ .
= -^—4-= 1250×10 4 =0.1250 m = 12.5 cm 2×10
Therefore, the wavelength of the radio waves is 12.5 cm.
See lessAnswer 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?
Ans (a). In a single slit diffraction experiment, 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 of the central diffraction band increases up to four times. Ans (b). The interference pattern in a double-slitRead more
Ans (a).
In a single slit diffraction experiment, 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 of the central diffraction band increases up to four times.
Ans (b).
The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The pattern is the result of the interference of the diffracted wave from each slit.
Ans (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. This is because light waves are diffracted from the edge of the circular obstacle, which interferes constructively at the centre of the shadow. This constructive interference produces a bright spot.
Ans (d).
Bending of waves by obstacles by a large angle is possible when the size of the obstacle is comparable to the wavelength of the waves.
On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.
Ans (e).
The justification is that in ordinary optical instruments, the size of the aperture involved is much larger than the wavelength of the light used.
See lessIn double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?
Wavelength of light used, λ = 6000 nm = 600 x 10-9 m Angular width of fringe, 0 = 0.1° = 0.1 x π/180 = 3.14 /1800 rad Angular width of a fringe is related to slit spacing (d) as: 0 = λ /d => d = λ/0 => d = (600 x 10-9) /(3.14/1800) = 3.44 x 10⁻4 m Therefore, the spacing between the slits isRead more
Wavelength of light used, λ = 6000 nm = 600 x 10-9 m
Angular width of fringe, 0 = 0.1° = 0.1 x π/180 = 3.14 /1800 rad
Angular width of a fringe is related to slit spacing (d) as: 0 = λ /d => d = λ/0
=> d = (600 x 10-9) /(3.14/1800) = 3.44 x 10⁻4 m
Therefore, the spacing between the slits is 3.44 x 10⁻4 m.
See lessFor 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?
No 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, sounRead more
No
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, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.
See lessLet us list some of the factors, which could possibly influence the speed of wave propagation: (i) Nature of the source. (ii) Direction of propagation. (iii) Motion of the source and/or observer. (iv) Wave length. (v) Intensity of the wave. On which of these factors, if any, does (a) The speed of light in vacuum, (b) The speed of light in a medium (say, glass or water), depend?
Ans (a). The speed of light in a vacuum i.e., 3 x 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum. Ans (b). Out of the listed factors, the speed of light inRead more
Ans (a).
The speed of light in a vacuum i.e., 3 x 108 m/s (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. Hence, the given factor does not affect the speed of light in a vacuum.
Ans (b).
Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that medium.
See lessExplain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
No; Wave theory Newton's corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the comRead more
No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.
Hence, we can write the expression:
c sin i = v sin r … (i)
Where, i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as: μ =v/c
Hence, equation (i) reduces to v/c = sini/sinr = μ … (2)
But, μ > 1
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v. The wave picture of light is consistent with the experimental results.
See lessThe 6563 Aº Ha line emitted by hydrogen in a star is found to be red shifted by 15 Aº. Estimate the speed with which the star is receding from the Earth.
Wavelength of Ha line emitted by hydrogen, λ = 6563 Aº = 6563 x 10⁻10 m. Star’s red-shift, (λ ' —λ ) = 15 Aº = 15 x 10-10 m Speed of light, c = 3 x 108 m/s Let the velocity of the star receding away from the Earth be v. The red shift is related with velocity as: λ ' —λ = v λ /c v = c/λ x (λ ' —λRead more
Wavelength of Ha line emitted by hydrogen, λ = 6563 Aº = 6563 x 10⁻10 m.
Star’s red-shift, (λ ‘ —λ ) = 15 Aº = 15 x 10-10 m
Speed of light, c = 3 x 108 m/s
Let the velocity of the star receding away from the Earth be v.
The red shift is related with velocity as:
λ ‘ —λ = v λ /c
v = c/λ x (λ ‘ —λ )
= (3 x 108 x 15 x 10-10 )/(6563 x 10⁻10) =6.87 x 10⁵
Therefore, the speed with which the star is receding away from the Earth is 6.87 x 10⁵ m/s.
See less