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LIGHT WAVES
QUESTIONS FOR SHORT ANSWER
1. Is the color of 620 nm light and 780 nm light same?
Is the color of 620 nm and 621 nm light same?
How many colors are there in white light?
ANSWER: Roughly the range of wavelengths from 620 nm to 780 nm is seen as red to human eyes. So the color of 620 nm and 780 nm can be broadly categorized as red but due to a large difference in the wavelengths both will not look same.
The range of wavelengths roughly from 590 nm to 620 nm is seen as orange to human eyes. But the color of 620 nm and 621 nm light may not be differentiated by the human eyes. Because the range is roughly divided, the 620 nm light and 621 nm light will appear same.
In the white light of Sun broadly seven colors are there. Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR), but a large number of shades of each of the colors can be distinguished by human eyes.
2. The wavelength of light in a medium is 𝞴 = 𝞴₀/µ, where 𝞴 is the wavelength in vacuum. A beam of red light (𝞴₀ = 720 nm) enters into the water. The wavelength in water is 𝞴 = 𝞴₀/µ = 540 nm. To a person under water does this light appears green?
ANSWER: No. It is the frequency of the visible light that determines the color perception to the human eyes. It is so because the energy of a photon depends on the frequency and this energy of the photon determines a particular color.
It can be understood in another way. When we see a light, it enters our eyes through eye-lens and passes through the fluid in the eyeballs before reaching the retina. Whether we see the light of a certain frequency in the air or in water it has to go through the fluid in the eyeballs. In that fluid, this frequency will have the same corresponding wavelength. Hence the same color.
3. Whether the diffraction effects from a slit will be more clearly visible or less clearly if the slit width is increased?
ANSWER: Since most of the diffracted light is distributed between an angle θ such that
sinθ = -𝞴/b to sinθ = 𝞴/b, where b is the slit width.
As b is increased, θ decreases. It means the cone of divergence narrows down and the diffraction effects will be less clearly visible.
4. If we put cardboard say (say 20 cm x 20 cm) between a light source and our eyes, we cant see the light. But when we put the same cardboard between a sound source and our ear, we hear the sound almost clearly. Explain.
ANSWER: Since the wavelength of a light wave is very small in comparison to sound wave, diffraction effects are not noticeable and the light almost travels in a straight line. Hence the cardboard is opaque and we cannot see the light.
In the case of sound waves the diffraction, reflection and refraction are prominent so it can be heard. Sound waves not only bend near the end of cardboard but it can vibrate the cardboard which in turn vibrate the air molecules on the other side to propagate the sound.
5. TV signals broadcast by Delhi Studio cannot be directly received at Patna which is almost 1000 km away. But the same signal goes some 36000 km away to a satellite, gets reflected and is then received at Patna. Explain.
ANSWER: TV signals are electromagnetic waves just like light waves. Hence it travels in a straight line. Due to the curvature of the earth, the TV signals broadcast by Delhi Studio get obstructed on the way to Patna. Hence it cannot be directly received at Patna.
Diagram for Q-5
But the same signal can be received by satellite and re-broadcasted which can be received at Patna. Because there is no obstruction in the straight path of the signals.
6. Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
ANSWER: Young's double slit experiment can be performed with sound waves also. To get a reasonable "fringe Pattern", the order of separation between the slits should be comparable to the wavelength of the sound.
The bright and dark fringes can be detected by measuring the intensity of the sound with an electronic detector or manually.
7. Is it necessary to have two waves of equal intensity to study interference pattern? Will there be an effect on clarity if the waves have unequal intensity?
ANSWER: No, it is not necessary to have two waves of equal intensity to study interference pattern. But with the waves of unequal intensity, the clarity of the interference pattern will be affected. Because the resultant field E at a point is given as
E² = E'² + E"² + 2E' E" cosẟ.
E will not be zero for any value of ẟ. The destructive (dark) interference will not be completely dark and the contrast will be low.
8. Can we conclude from the interference phenomenon whether light is a transverse wave or longitudinal wave?
ANSWER: No. Interference phenomenon only indicates the wave nature of light, it can not be concluded whether it is a transverse wave or longitudinal wave because it occurs in both types of waves.
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?
ANSWER: Interference occurs only with coherent sources which have an initial phase difference that does not change with time and have the same frequency. Two candles are not coherent sources hence we do not have an interference pattern.
When the candles are replaced by laser sources they can produce interference pattern provided that they fulfill the conditions of coherent sources.
10. If the separation between the slits in Young's double slit experiment is increased, what happens to the fringe width? If the separation is increased too much, will the fringe pattern remain detectable?
ANSWER: In Young's double slit experiment the fringe width is given as
w = D𝜆/d, where d is the separation between the slits. Clearly, if d is increased the fringe width decreases.
If the separation d is increased too much the fringe width will be very small. Now the maxima and minima will be so closely spaced that it will look uniform intensity pattern.
11. Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing 𝞴 = 400 nm). Describe the nature of the fringe pattern observed.
ANSWER: Suppose the white light comes from the same source. The white light is a combination of a range of colors having different frequencies. The fringe patterns are found when monochromatic light is used. Since here one of the slit is covered by a violet filter, only violet light of 𝝀 = 400 nm will pass through it. This violet light will interfere with the violet light of the frequency 𝝀 = 400 nm (the constituent of the white light) coming from another slit and form fringe patterns. The maxima will be of violet color while minima will not be completely dark because other frequencies of white light will fall here.
1. Is the color of 620 nm light and 780 nm light same?
Is the color of 620 nm and 621 nm light same?
How many colors are there in white light?
ANSWER: Roughly the range of wavelengths from 620 nm to 780 nm is seen as red to human eyes. So the color of 620 nm and 780 nm can be broadly categorized as red but due to a large difference in the wavelengths both will not look same.
The range of wavelengths roughly from 590 nm to 620 nm is seen as orange to human eyes. But the color of 620 nm and 621 nm light may not be differentiated by the human eyes. Because the range is roughly divided, the 620 nm light and 621 nm light will appear same.
In the white light of Sun broadly seven colors are there. Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR), but a large number of shades of each of the colors can be distinguished by human eyes.
2. The wavelength of light in a medium is 𝞴 = 𝞴₀/µ, where 𝞴 is the wavelength in vacuum. A beam of red light (𝞴₀ = 720 nm) enters into the water. The wavelength in water is 𝞴 = 𝞴₀/µ = 540 nm. To a person under water does this light appears green?
ANSWER: No. It is the frequency of the visible light that determines the color perception to the human eyes. It is so because the energy of a photon depends on the frequency and this energy of the photon determines a particular color.
It can be understood in another way. When we see a light, it enters our eyes through eye-lens and passes through the fluid in the eyeballs before reaching the retina. Whether we see the light of a certain frequency in the air or in water it has to go through the fluid in the eyeballs. In that fluid, this frequency will have the same corresponding wavelength. Hence the same color.
3. Whether the diffraction effects from a slit will be more clearly visible or less clearly if the slit width is increased?
ANSWER: Since most of the diffracted light is distributed between an angle θ such that
sinθ = -𝞴/b to sinθ = 𝞴/b, where b is the slit width.
As b is increased, θ decreases. It means the cone of divergence narrows down and the diffraction effects will be less clearly visible.
4. If we put cardboard say (say 20 cm x 20 cm) between a light source and our eyes, we cant see the light. But when we put the same cardboard between a sound source and our ear, we hear the sound almost clearly. Explain.
ANSWER: Since the wavelength of a light wave is very small in comparison to sound wave, diffraction effects are not noticeable and the light almost travels in a straight line. Hence the cardboard is opaque and we cannot see the light.
In the case of sound waves the diffraction, reflection and refraction are prominent so it can be heard. Sound waves not only bend near the end of cardboard but it can vibrate the cardboard which in turn vibrate the air molecules on the other side to propagate the sound.
5. TV signals broadcast by Delhi Studio cannot be directly received at Patna which is almost 1000 km away. But the same signal goes some 36000 km away to a satellite, gets reflected and is then received at Patna. Explain.
ANSWER: TV signals are electromagnetic waves just like light waves. Hence it travels in a straight line. Due to the curvature of the earth, the TV signals broadcast by Delhi Studio get obstructed on the way to Patna. Hence it cannot be directly received at Patna.
But the same signal can be received by satellite and re-broadcasted which can be received at Patna. Because there is no obstruction in the straight path of the signals.
6. Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
ANSWER: Young's double slit experiment can be performed with sound waves also. To get a reasonable "fringe Pattern", the order of separation between the slits should be comparable to the wavelength of the sound.
The bright and dark fringes can be detected by measuring the intensity of the sound with an electronic detector or manually.
7. Is it necessary to have two waves of equal intensity to study interference pattern? Will there be an effect on clarity if the waves have unequal intensity?
ANSWER: No, it is not necessary to have two waves of equal intensity to study interference pattern. But with the waves of unequal intensity, the clarity of the interference pattern will be affected. Because the resultant field E at a point is given as
E² = E'² + E"² + 2E' E" cosẟ.
E will not be zero for any value of ẟ. The destructive (dark) interference will not be completely dark and the contrast will be low.
8. Can we conclude from the interference phenomenon whether light is a transverse wave or longitudinal wave?
ANSWER: No. Interference phenomenon only indicates the wave nature of light, it can not be concluded whether it is a transverse wave or longitudinal wave because it occurs in both types of waves.
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?
ANSWER: Interference occurs only with coherent sources which have an initial phase difference that does not change with time and have the same frequency. Two candles are not coherent sources hence we do not have an interference pattern.
When the candles are replaced by laser sources they can produce interference pattern provided that they fulfill the conditions of coherent sources.
10. If the separation between the slits in Young's double slit experiment is increased, what happens to the fringe width? If the separation is increased too much, will the fringe pattern remain detectable?
ANSWER: In Young's double slit experiment the fringe width is given as
w = D𝜆/d, where d is the separation between the slits. Clearly, if d is increased the fringe width decreases.
If the separation d is increased too much the fringe width will be very small. Now the maxima and minima will be so closely spaced that it will look uniform intensity pattern.
11. Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing 𝞴 = 400 nm). Describe the nature of the fringe pattern observed.
ANSWER: Suppose the white light comes from the same source. The white light is a combination of a range of colors having different frequencies. The fringe patterns are found when monochromatic light is used. Since here one of the slit is covered by a violet filter, only violet light of 𝝀 = 400 nm will pass through it. This violet light will interfere with the violet light of the frequency 𝝀 = 400 nm (the constituent of the white light) coming from another slit and form fringe patterns. The maxima will be of violet color while minima will not be completely dark because other frequencies of white light will fall here.
Is the color of 620 nm and 621 nm light same?
How many colors are there in white light?
ANSWER: Roughly the range of wavelengths from 620 nm to 780 nm is seen as red to human eyes. So the color of 620 nm and 780 nm can be broadly categorized as red but due to a large difference in the wavelengths both will not look same.
The range of wavelengths roughly from 590 nm to 620 nm is seen as orange to human eyes. But the color of 620 nm and 621 nm light may not be differentiated by the human eyes. Because the range is roughly divided, the 620 nm light and 621 nm light will appear same.
In the white light of Sun broadly seven colors are there. Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR), but a large number of shades of each of the colors can be distinguished by human eyes.
2. The wavelength of light in a medium is 𝞴 = 𝞴₀/µ, where 𝞴 is the wavelength in vacuum. A beam of red light (𝞴₀ = 720 nm) enters into the water. The wavelength in water is 𝞴 = 𝞴₀/µ = 540 nm. To a person under water does this light appears green?
ANSWER: No. It is the frequency of the visible light that determines the color perception to the human eyes. It is so because the energy of a photon depends on the frequency and this energy of the photon determines a particular color.
It can be understood in another way. When we see a light, it enters our eyes through eye-lens and passes through the fluid in the eyeballs before reaching the retina. Whether we see the light of a certain frequency in the air or in water it has to go through the fluid in the eyeballs. In that fluid, this frequency will have the same corresponding wavelength. Hence the same color.
3. Whether the diffraction effects from a slit will be more clearly visible or less clearly if the slit width is increased?
ANSWER: Since most of the diffracted light is distributed between an angle θ such that
sinθ = -𝞴/b to sinθ = 𝞴/b, where b is the slit width.
As b is increased, θ decreases. It means the cone of divergence narrows down and the diffraction effects will be less clearly visible.
4. If we put cardboard say (say 20 cm x 20 cm) between a light source and our eyes, we cant see the light. But when we put the same cardboard between a sound source and our ear, we hear the sound almost clearly. Explain.
ANSWER: Since the wavelength of a light wave is very small in comparison to sound wave, diffraction effects are not noticeable and the light almost travels in a straight line. Hence the cardboard is opaque and we cannot see the light.
In the case of sound waves the diffraction, reflection and refraction are prominent so it can be heard. Sound waves not only bend near the end of cardboard but it can vibrate the cardboard which in turn vibrate the air molecules on the other side to propagate the sound.
5. TV signals broadcast by Delhi Studio cannot be directly received at Patna which is almost 1000 km away. But the same signal goes some 36000 km away to a satellite, gets reflected and is then received at Patna. Explain.
ANSWER: TV signals are electromagnetic waves just like light waves. Hence it travels in a straight line. Due to the curvature of the earth, the TV signals broadcast by Delhi Studio get obstructed on the way to Patna. Hence it cannot be directly received at Patna.
 |
| Diagram for Q-5 |
But the same signal can be received by satellite and re-broadcasted which can be received at Patna. Because there is no obstruction in the straight path of the signals.
6. Can we perform Young's double slit experiment with sound waves? To get a reasonable "fringe pattern", what should be the order of separation between the slits? How can the bright fringes and the dark fringes be detected in this case?
ANSWER: Young's double slit experiment can be performed with sound waves also. To get a reasonable "fringe Pattern", the order of separation between the slits should be comparable to the wavelength of the sound.
The bright and dark fringes can be detected by measuring the intensity of the sound with an electronic detector or manually.
7. Is it necessary to have two waves of equal intensity to study interference pattern? Will there be an effect on clarity if the waves have unequal intensity?
ANSWER: No, it is not necessary to have two waves of equal intensity to study interference pattern. But with the waves of unequal intensity, the clarity of the interference pattern will be affected. Because the resultant field E at a point is given as
E² = E'² + E"² + 2E' E" cosẟ.
E will not be zero for any value of ẟ. The destructive (dark) interference will not be completely dark and the contrast will be low.
8. Can we conclude from the interference phenomenon whether light is a transverse wave or longitudinal wave?
ANSWER: No. Interference phenomenon only indicates the wave nature of light, it can not be concluded whether it is a transverse wave or longitudinal wave because it occurs in both types of waves.
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?
ANSWER: Interference occurs only with coherent sources which have an initial phase difference that does not change with time and have the same frequency. Two candles are not coherent sources hence we do not have an interference pattern.
When the candles are replaced by laser sources they can produce interference pattern provided that they fulfill the conditions of coherent sources.
10. If the separation between the slits in Young's double slit experiment is increased, what happens to the fringe width? If the separation is increased too much, will the fringe pattern remain detectable?
ANSWER: In Young's double slit experiment the fringe width is given as
w = D𝜆/d, where d is the separation between the slits. Clearly, if d is increased the fringe width decreases.
If the separation d is increased too much the fringe width will be very small. Now the maxima and minima will be so closely spaced that it will look uniform intensity pattern.
11. Suppose white light falls on a double slit but one slit is covered by a violet filter (allowing 𝞴 = 400 nm). Describe the nature of the fringe pattern observed.
ANSWER: Suppose the white light comes from the same source. The white light is a combination of a range of colors having different frequencies. The fringe patterns are found when monochromatic light is used. Since here one of the slit is covered by a violet filter, only violet light of 𝝀 = 400 nm will pass through it. This violet light will interfere with the violet light of the frequency 𝝀 = 400 nm (the constituent of the white light) coming from another slit and form fringe patterns. The maxima will be of violet color while minima will not be completely dark because other frequencies of white light will fall here.
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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