Wave Optics - Important Questions and Answers
Multiple Choice Questions (MCQs):
1. Which of the following phenomena is explained by Huygens' principle?
a) Photoelectric effect
b) Reflection and refraction
c) Compton effect
d) Raman effect
a) Photoelectric effect
b) Reflection and refraction
c) Compton effect
d) Raman effect
Answer: b) Reflection and refraction
Concept: Huygens' principle is used to explain the wave nature of light and phenomena such as reflection and refraction.
Concept: Huygens' principle is used to explain the wave nature of light and phenomena such as reflection and refraction.
2. In Young’s double slit experiment, the fringe width increases if:
a) The wavelength of light decreases
b) The distance between the slits increases
c) The distance between the screen and the slits decreases
d) The wavelength of light increases
a) The wavelength of light decreases
b) The distance between the slits increases
c) The distance between the screen and the slits decreases
d) The wavelength of light increases
Answer: d) The wavelength of light increases
Concept: Fringe width \( (\beta) \) is given by \( \beta = \frac{\lambda D}{d} \), where \( \lambda \) is the wavelength, \( D \) is the distance to the screen, and \( d \) is the distance between the slits.
Concept: Fringe width \( (\beta) \) is given by \( \beta = \frac{\lambda D}{d} \), where \( \lambda \) is the wavelength, \( D \) is the distance to the screen, and \( d \) is the distance between the slits.
3. The ratio of intensities of two waves interfering at a point is 9:4. The ratio of the amplitudes of the two waves is:
a) 3:2
b) 2:3
c) 9:4
d) 4:9
a) 3:2
b) 2:3
c) 9:4
d) 4:9
Answer: a) 3:2
Concept: Intensity \( (I) \) is proportional to the square of the amplitude \( (A) \), so \( I_1/I_2 = (A_1/A_2)^2 \). Given \( I_1/I_2 = 9/4 \), \( A_1/A_2 = 3/2 \).
Concept: Intensity \( (I) \) is proportional to the square of the amplitude \( (A) \), so \( I_1/I_2 = (A_1/A_2)^2 \). Given \( I_1/I_2 = 9/4 \), \( A_1/A_2 = 3/2 \).
4. In Young's double slit experiment, coherent sources are required for sustained interference because:
a) They have different wavelengths
b) They maintain a constant phase difference
c) They have different frequencies
d) They produce high intensity light
a) They have different wavelengths
b) They maintain a constant phase difference
c) They have different frequencies
d) They produce high intensity light
Answer: b) They maintain a constant phase difference
Concept: Coherent sources maintain a constant phase difference, which is necessary for sustained interference patterns.
Concept: Coherent sources maintain a constant phase difference, which is necessary for sustained interference patterns.
5. The angular width of the central maximum in a single slit diffraction pattern is:
a) Directly proportional to the slit width
b) Inversely proportional to the slit width
c) Directly proportional to the wavelength of light
d) Inversely proportional to the wavelength of light
a) Directly proportional to the slit width
b) Inversely proportional to the slit width
c) Directly proportional to the wavelength of light
d) Inversely proportional to the wavelength of light
Answer: b) Inversely proportional to the slit width
Concept: The angular width \( \theta \) is given by \( \theta = \frac{\lambda}{a} \), where \( \lambda \) is the wavelength and \( a \) is the slit width.
Concept: The angular width \( \theta \) is given by \( \theta = \frac{\lambda}{a} \), where \( \lambda \) is the wavelength and \( a \) is the slit width.
6. The condition for sustained interference of light waves is:
a) The sources must be incoherent
b) The sources must be coherent
c) The sources must be of different wavelengths
d) The sources must have different frequencies
a) The sources must be incoherent
b) The sources must be coherent
c) The sources must be of different wavelengths
d) The sources must have different frequencies
Answer: b) The sources must be coherent
Concept: Coherent sources are necessary for sustained interference because they maintain a constant phase relationship.
Concept: Coherent sources are necessary for sustained interference because they maintain a constant phase relationship.
7. In Young’s double slit experiment, the distance between the slits is doubled. The fringe width will:
a) Remain the same
b) Be doubled
c) Be halved
d) Become four times
a) Remain the same
b) Be doubled
c) Be halved
d) Become four times
Answer: c) Be halved
Concept: Fringe width \( (\beta) \) is inversely proportional to the distance between the slits \( (d) \), so doubling \( d \) will halve \( \beta \).
Concept: Fringe width \( (\beta) \) is inversely proportional to the distance between the slits \( (d) \), so doubling \( d \) will halve \( \beta \).
8. The phenomenon where light bends around the corners of an obstacle is called:
a) Interference
b) Diffraction
c) Reflection
d) Refraction
a) Interference
b) Diffraction
c) Reflection
d) Refraction
Answer: b) Diffraction
Concept: Diffraction is the bending of light waves around the edges of an obstacle.
Concept: Diffraction is the bending of light waves around the edges of an obstacle.
9. The wavelength of light used in Young's double slit experiment is 600 nm. If the distance between the slits is 0.3 mm, and the distance between the slits and the screen is 1.5 m, the fringe width is:
a) 3 mm
b) 2 mm
c) 1 mm
d) 0.5 mm
a) 3 mm
b) 2 mm
c) 1 mm
d) 0.5 mm
Answer: b) 2 mm
Concept: Fringe width \( \beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.5}{0.3 \times 10^{-3}} = 3 \times 10^{-3} \) m or 3 mm.
Concept: Fringe width \( \beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.5}{0.3 \times 10^{-3}} = 3 \times 10^{-3} \) m or 3 mm.
10. In a single slit diffraction experiment, if the width of the slit is halved, the width of the central maximum will:
a) Be halved
b) Be doubled
c) Remain the same
d) Become four times
a) Be halved
b) Be doubled
c) Remain the same
d) Become four times
Answer: b) Be doubled
Concept: The width of the central maximum is inversely proportional to the slit width, so halving the slit width will double the width of the central maximum.
Concept: The width of the central maximum is inversely proportional to the slit width, so halving the slit width will double the width of the central maximum.
Subjective Questions:
1. Explain the concept of wavefront. Using Huygens’ principle, derive the laws of reflection.
Answer:
Concept: A wavefront is the locus of points having the same phase. According to Huygens' principle, each point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the tangential surface to these secondary wavelets.
Derivation:
For reflection, consider a wavefront AB incident at an angle on a reflecting surface. According to Huygens' principle, each point on AB acts as a source of secondary wavelets. The reflected wavefront CD is formed such that it is tangential to these wavelets. By geometrical construction, it can be shown that the angle of incidence equals the angle of reflection.
Concept: A wavefront is the locus of points having the same phase. According to Huygens' principle, each point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the tangential surface to these secondary wavelets.
Derivation:
For reflection, consider a wavefront AB incident at an angle on a reflecting surface. According to Huygens' principle, each point on AB acts as a source of secondary wavelets. The reflected wavefront CD is formed such that it is tangential to these wavelets. By geometrical construction, it can be shown that the angle of incidence equals the angle of reflection.
2. Using Huygens’ principle, explain the refraction of a plane wavefront at a plane surface.
Answer:
Concept: Huygens' principle states that each point on a wavefront is a source of secondary wavelets. The new wavefront is the tangential surface to these wavelets.
Explanation:
When a plane wavefront AB enters a denser medium, the speed of light decreases. The secondary wavelets in the denser medium will have a smaller radius than in the rarer medium. The new wavefront CD in the denser medium is formed such that it is tangential to these wavelets. By geometrical construction, it can be shown that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, which is Snell's law.
Concept: Huygens' principle states that each point on a wavefront is a source of secondary wavelets. The new wavefront is the tangential surface to these wavelets.
Explanation:
When a plane wavefront AB enters a denser medium, the speed of light decreases. The secondary wavelets in the denser medium will have a smaller radius than in the rarer medium. The new wavefront CD in the denser medium is formed such that it is tangential to these wavelets. By geometrical construction, it can be shown that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, which is Snell's law.
3. State Young’s double slit experiment and provide the expression for fringe width.
Answer:
Concept: Young's double slit experiment demonstrates the wave nature of light through interference.
Experiment:
Light from a single source passes through two slits, producing two coherent sources. These coherent sources interfere, forming an interference pattern on a screen. The bright and dark fringes result from constructive and destructive interference.
Fringe Width:
The fringe width \( \beta \) is given by \( \beta = \frac{\lambda D}{d} \), where \( \lambda \) is the wavelength of light, \( D \) is the distance between the slits and the screen, and \( d \) is the distance between the slits.
Concept: Young's double slit experiment demonstrates the wave nature of light through interference.
Experiment:
Light from a single source passes through two slits, producing two coherent sources. These coherent sources interfere, forming an interference pattern on a screen. The bright and dark fringes result from constructive and destructive interference.
Fringe Width:
The fringe width \( \beta \) is given by \( \beta = \frac{\lambda D}{d} \), where \( \lambda \) is the wavelength of light, \( D \) is the distance between the slits and the screen, and \( d \) is the distance between the slits.
4. What is meant by coherent sources? Why are coherent sources necessary for sustained interference?
Answer:
Concept: Coherent sources are light sources that maintain a constant phase difference and have the same frequency.
Explanation:
Coherent sources are necessary for sustained interference because only coherent sources maintain a stable and consistent phase relationship over time. This stable phase difference results in constructive and destructive interference patterns that do not vary with time.
Concept: Coherent sources are light sources that maintain a constant phase difference and have the same frequency.
Explanation:
Coherent sources are necessary for sustained interference because only coherent sources maintain a stable and consistent phase relationship over time. This stable phase difference results in constructive and destructive interference patterns that do not vary with time.
5. Describe the phenomenon of diffraction of light through a single slit. Provide a qualitative treatment of the width of the central maximum.
Answer:
Concept: Diffraction is the bending of light waves around the edges of an obstacle or through an aperture.
Explanation:
In a single slit diffraction experiment, light passing through a slit produces a diffraction pattern of dark and bright regions on a screen. The central maximum is the brightest and widest part of the pattern.
Width of Central Maximum:
The width of the central maximum is given by \( \theta = \frac{\lambda}{a} \), where \( \lambda \) is the wavelength of light and \( a \) is the width of the slit. This means the central maximum becomes wider as the slit width decreases.
Concept: Diffraction is the bending of light waves around the edges of an obstacle or through an aperture.
Explanation:
In a single slit diffraction experiment, light passing through a slit produces a diffraction pattern of dark and bright regions on a screen. The central maximum is the brightest and widest part of the pattern.
Width of Central Maximum:
The width of the central maximum is given by \( \theta = \frac{\lambda}{a} \), where \( \lambda \) is the wavelength of light and \( a \) is the width of the slit. This means the central maximum becomes wider as the slit width decreases.
6. In Young’s double slit experiment, how does the fringe width change if the whole apparatus is immersed in water? Explain with reasons.
Answer:
Concept: The refractive index of water affects the wavelength of light.
Explanation:
The wavelength of light in water is given by \( \lambda' = \frac{\lambda}{n} \), where \( n \) is the refractive index of water. Since \( \beta = \frac{\lambda D}{d} \), the fringe width \( \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{nd} \). Therefore, the fringe width decreases by a factor of \( n \).
Concept: The refractive index of water affects the wavelength of light.
Explanation:
The wavelength of light in water is given by \( \lambda' = \frac{\lambda}{n} \), where \( n \) is the refractive index of water. Since \( \beta = \frac{\lambda D}{d} \), the fringe width \( \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{nd} \). Therefore, the fringe width decreases by a factor of \( n \).
7. Explain how interference patterns are formed in Young’s double slit experiment. What conditions are necessary for observing these patterns?
Answer:
Concept: Interference patterns result from the superposition of two coherent light waves.
Explanation:
In Young's double slit experiment, light from two slits acts as two coherent sources. These waves superpose, resulting in constructive interference (bright fringes) and destructive interference (dark fringes).
Conditions:
- The light sources must be coherent.
- The light waves should have the same frequency and nearly equal amplitudes.
- The slits should be narrow and closely spaced.
Concept: Interference patterns result from the superposition of two coherent light waves.
Explanation:
In Young's double slit experiment, light from two slits acts as two coherent sources. These waves superpose, resulting in constructive interference (bright fringes) and destructive interference (dark fringes).
Conditions:
- The light sources must be coherent.
- The light waves should have the same frequency and nearly equal amplitudes.
- The slits should be narrow and closely spaced.
8. A light wave traveling in air is incident on a glass slab at an angle of 30°. Using Huygens’ principle, explain the change in the wavefront as it enters the glass slab.
Answer:
Concept: Huygens' principle explains the behavior of wavefronts during refraction.
Explanation:
As a wavefront enters the glass slab, each point on the wavefront acts as a source of secondary wavelets. The speed of light in glass is less than in air, so the secondary wavelets in glass have a smaller radius. The wavefront bends towards the normal as it enters the glass. According to Snell's law, \( \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \), where \( i \) is the angle of incidence, \( r \) is the angle of refraction, \( v_1 \) is the speed of light in air, and \( v_2 \) is the speed of light in glass.
Concept: Huygens' principle explains the behavior of wavefronts during refraction.
Explanation:
As a wavefront enters the glass slab, each point on the wavefront acts as a source of secondary wavelets. The speed of light in glass is less than in air, so the secondary wavelets in glass have a smaller radius. The wavefront bends towards the normal as it enters the glass. According to Snell's law, \( \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \), where \( i \) is the angle of incidence, \( r \) is the angle of refraction, \( v_1 \) is the speed of light in air, and \( v_2 \) is the speed of light in glass.
9. Derive the relationship between the angle of incidence and the angle of refraction using Huygens’ principle.
Answer:
Concept: Huygens' principle and Snell's law.
Derivation:
Consider a wavefront AB incident at an angle \( i \) on a refracting surface. According to Huygens' principle, each point on AB acts as a source of secondary wavelets. The wavelets in the denser medium have a smaller radius. The new wavefront CD in the denser medium is formed such that it is tangential to these wavelets. By geometry and using the definition of sine, it can be shown that \( \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \), where \( v_1 \) and \( v_2 \) are the speeds of light in the two media.
Concept: Huygens' principle and Snell's law.
Derivation:
Consider a wavefront AB incident at an angle \( i \) on a refracting surface. According to Huygens' principle, each point on AB acts as a source of secondary wavelets. The wavelets in the denser medium have a smaller radius. The new wavefront CD in the denser medium is formed such that it is tangential to these wavelets. By geometry and using the definition of sine, it can be shown that \( \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \), where \( v_1 \) and \( v_2 \) are the speeds of light in the two media.
10. Calculate the fringe width in Young’s double slit experiment if the wavelength of light used is 500 nm, the distance between the slits is 0.25 mm, and the distance between the slits and the screen is 1 m.
Answer:
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
Given \( \lambda = 500 \times 10^{-9} \) m, \( d = 0.25 \times 10^{-3} \) m, \( D = 1 \) m
\( \beta = \frac{500 \times 10^{-9} \times 1}{0.25 \times 10^{-3}} = 2 \times 10^{-3} \) m or 2 mm
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
Given \( \lambda = 500 \times 10^{-9} \) m, \( d = 0.25 \times 10^{-3} \) m, \( D = 1 \) m
\( \beta = \frac{500 \times 10^{-9} \times 1}{0.25 \times 10^{-3}} = 2 \times 10^{-3} \) m or 2 mm
11. Light of wavelength 600 nm is used in a double slit experiment. The slits are 0.5 mm apart, and the screen is 1.5 m away from the slits. Calculate the distance between the central bright fringe and the first bright fringe.
Answer:
Concept: Position of fringes in Young's double slit experiment.
Formula: \( y_n = n \frac{\lambda D}{d} \)
Calculation:
For the first bright fringe \( (n = 1) \)
Given \( \lambda = 600 \times 10^{-9} \) m, \( d = 0.5 \times 10^{-3} \) m, \( D = 1.5 \) m
\( y_1 = 1 \times \frac{600 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}} = 1.8 \times 10^{-3} \) m or 1.8 mm
Concept: Position of fringes in Young's double slit experiment.
Formula: \( y_n = n \frac{\lambda D}{d} \)
Calculation:
For the first bright fringe \( (n = 1) \)
Given \( \lambda = 600 \times 10^{-9} \) m, \( d = 0.5 \times 10^{-3} \) m, \( D = 1.5 \) m
\( y_1 = 1 \times \frac{600 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}} = 1.8 \times 10^{-3} \) m or 1.8 mm
12. A beam of light of wavelength 450 nm is incident on a slit of width 0.3 mm. Calculate the angular width of the central maximum in the resulting diffraction pattern.
Answer:
Concept: Angular width of central maximum in single slit diffraction.
Formula: \( \theta = \frac{\lambda}{a} \)
Calculation:
Given \( \lambda = 450 \times 10^{-9} \) m, \( a = 0.3 \times 10^{-3} \) m
\( \theta = \frac{450 \times 10^{-9}}{0.3 \times 10^{-3}} = 1.5 \times 10^{-3} \) rad or 0.0015 rad
Concept: Angular width of central maximum in single slit diffraction.
Formula: \( \theta = \frac{\lambda}{a} \)
Calculation:
Given \( \lambda = 450 \times 10^{-9} \) m, \( a = 0.3 \times 10^{-3} \) m
\( \theta = \frac{450 \times 10^{-9}}{0.3 \times 10^{-3}} = 1.5 \times 10^{-3} \) rad or 0.0015 rad
13. In a double slit experiment, if the wavelength of light is 700 nm and the slits are separated by 0.25 mm, calculate the fringe width when the screen is 2 m away from the slits.
Answer:
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
Given \( \lambda = 700 \times 10^{-9} \) m, \( d = 0.25 \times 10^{-3} \) m, \( D = 2 \) m
\( \beta = \frac{700 \times 10^{-9} \times 2}{0.25 \times 10^{-3}} = 5.6 \times 10^{-3} \) m or 5.6 mm
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
Given \( \lambda = 700 \times 10^{-9} \) m, \( d = 0.25 \times 10^{-3} \) m, \( D = 2 \) m
\( \beta = \frac{700 \times 10^{-9} \times 2}{0.25 \times 10^{-3}} = 5.6 \times 10^{-3} \) m or 5.6 mm
14. A slit of width 0.5 mm is illuminated with light of wavelength 400 nm. Calculate the width of the central maximum on a screen 1 m away.
Answer:
Concept: Width of central maximum in single slit diffraction.
Formula: Width of central maximum \( = 2\theta \) where \( \theta = \frac{\lambda}{a} \)
Calculation:
Given \( \lambda = 400 \times 10^{-9} \) m, \( a = 0.5 \times 10^{-3} \) m, \( D = 1 \) m
\( \theta = \frac{400 \times 10^{-9}}{0.5 \times 10^{-3}} = 0.8 \times 10^{-3} \) rad
Width of central maximum \( = 2 \theta D = 2 \times 0.8 \times 10^{-3} \times 1 = 1.6 \times 10^{-3} \) m or 1.6 mm
Concept: Width of central maximum in single slit diffraction.
Formula: Width of central maximum \( = 2\theta \) where \( \theta = \frac{\lambda}{a} \)
Calculation:
Given \( \lambda = 400 \times 10^{-9} \) m, \( a = 0.5 \times 10^{-3} \) m, \( D = 1 \) m
\( \theta = \frac{400 \times 10^{-9}}{0.5 \times 10^{-3}} = 0.8 \times 10^{-3} \) rad
Width of central maximum \( = 2 \theta D = 2 \times 0.8 \times 10^{-3} \times 1 = 1.6 \times 10^{-3} \) m or 1.6 mm
15. If in Young’s double slit experiment, the distance between the slits and the screen is doubled and the wavelength of the light is halved, what will be the effect on the fringe width?
Answer:
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
If \( D \) is doubled and \( \lambda \) is halved, the new fringe width \( \beta' = \frac{(\lambda/2) \times (2D)}{d} = \frac{\lambda D}{d} = \beta \). Thus, the fringe width remains unchanged.
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
If \( D \) is doubled and \( \lambda \) is halved, the new fringe width \( \beta' = \frac{(\lambda/2) \times (2D)}{d} = \frac{\lambda D}{d} = \beta \). Thus, the fringe width remains unchanged.
16. A parallel beam of light of wavelength 500 nm is incident on a single slit of width 0.2 mm. Calculate the angular width of the central maximum.
Answer:
Concept: Angular width of central maximum in single slit diffraction.
Formula: \( \theta = \frac{\lambda}{a} \)
Calculation:
Given \( \lambda = 500 \times 10^{-9} \) m, \( a = 0.2 \times 10^{-3} \) m
\( \theta = \frac{500 \times 10^{-9}}{0.2 \times 10^{-3}} = 2.5 \times 10^{-3} \) rad or 0.0025 rad
Concept: Angular width of central maximum in single slit diffraction.
Formula: \( \theta = \frac{\lambda}{a} \)
Calculation:
Given \( \lambda = 500 \times 10^{-9} \) m, \( a = 0.2 \times 10^{-3} \) m
\( \theta = \frac{500 \times 10^{-9}}{0.2 \times 10^{-3}} = 2.5 \times 10^{-3} \) rad or 0.0025 rad
17. Two slits in Young's double slit experiment have widths in the ratio 1:4. What will be the ratio of the intensities of the bright fringes produced by these slits?
Answer:
Concept: Intensity in Young's double slit experiment.
Formula: Intensity \( I \) is proportional to the square of the amplitude \( A \).
Calculation:
If the widths are in the ratio 1:4, the amplitudes are in the ratio 1:4. Therefore, the intensities will be in the ratio \( (1^2):(4^2) = 1:16 \).
Concept: Intensity in Young's double slit experiment.
Formula: Intensity \( I \) is proportional to the square of the amplitude \( A \).
Calculation:
If the widths are in the ratio 1:4, the amplitudes are in the ratio 1:4. Therefore, the intensities will be in the ratio \( (1^2):(4^2) = 1:16 \).
18. In a Young's double slit experiment, the slits are separated by 0.4 mm and the screen is placed 1 m away. If the distance between the central bright fringe and the first bright fringe is 2.5 mm, calculate the wavelength of the light used.
Answer:
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
Given \( d = 0.4 \times 10^{-3} \) m, \( D = 1 \) m, \( \beta = 2.5 \times 10^{-3} \) m
\( \lambda = \frac{\beta d}{D} = \frac{2.5 \times 10^{-3} \times 0.4 \times 10^{-3}}{1} = 1 \times 10^{-6} \) m or 1000 nm
Concept: Fringe width in Young's double slit experiment.
Formula: \( \beta = \frac{\lambda D}{d} \)
Calculation:
Given \( d = 0.4 \times 10^{-3} \) m, \( D = 1 \) m, \( \beta = 2.5 \times 10^{-3} \) m
\( \lambda = \frac{\beta d}{D} = \frac{2.5 \times 10^{-3} \times 0.4 \times 10^{-3}}{1} = 1 \times 10^{-6} \) m or 1000 nm
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