The resolving power of a telescope can be increased by increasing:
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Wave Optics question types
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01
M.C.Q (1 Marks)
177 Q→02Assertion (A) & Reason (B) MCQ
20 Q→031 Marks Question
25 Q→042 Marks Questions
66 Q→053 Marks Question
55 Q→064 Marks Questions
13 Q→075 Marks Questions
32 Q→08Fill in the blanks.
1 Q→Sample Questions
Wave Optics questions
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Sound waves in air cannot be polarized because:
A wavefront is an imaginary surface where:
According to Maxwell , most of the optical properties of light depend on:
To increase both the resolving power and magnifying power of a telescope:
For question two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
Reason (R): Wave diffracted from the edges of circular obstacle interfere constructively at the centre of the shadow resulting in the formation of bright spot.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is NOT the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason (R): Wave diffracted from the edges of circular obstacle interfere constructively at the centre of the shadow resulting in the formation of bright spot.
For question two statements are given-one labelled Assertion $(A) $ and the other labelled Reason $(R).$ Select the correct answer to these questions from the codes $(a), (b), (c)$ and $(d)$ as given below. Assertion $(A)$: Two point coherent sources of light $S_1$ and $S_2$ are placed on a line as shown. $ P$ and $Q$ are two points on that line. If at point $P$ maximum intensity is observed then maximum intensity should also be observed at $Q$.
Reason $(R)$: In the figure of assertion the distance $|S_1P - S_2P|$ is equal to distance $|S_2Q- S_1Q|.$
Reason $(R)$: In the figure of assertion the distance $|S_1P - S_2P|$ is equal to distance $|S_2Q- S_1Q|.$
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A$.
- ✓Both $A$ and $R$ are true but $R $ is $\text{NOT}$ the correct explanation of $A$.
- C$A$ is true but $R$ is false.
- D$A$ is false and $R$ is also false.
Answer: B.
View full solution →For question two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
Reason (R): Destructive interference occurs at the centre of the shadow.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is NOT the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason (R): Destructive interference occurs at the centre of the shadow.
For question two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
Reason (R): In interference fringe width is independent of position of the fringe.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is NOT the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason (R): In interference fringe width is independent of position of the fringe.
For question two statements are given-one labelled Assertion $(A)$ and the other labelled Reason $(R)$. Select the correct answer to these questions from the codes $(a), (b), (c)$ and $(d)$ as given below.
Assertion $(A)$: In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, a dark fringe is observed. Then the wavelength of wave is proportional to square of distance of two slits.
Reason $(R)$ : For a dark fringe intensity is zero.
Assertion $(A)$: In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, a dark fringe is observed. Then the wavelength of wave is proportional to square of distance of two slits.
Reason $(R)$ : For a dark fringe intensity is zero.
- ABoth $A $ and $R$ are true and $R$ is the correct explanation of $A$.
- BBoth $A$ and $R$ are true but $R $ is $\text{NOT}$ the correct explanation of $A$.
- C$A$ is true but $R$ is false.
- ✓$A$ is false and $R$ is also false.
Answer: D.
View full solution →Answer the following question:
In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
Answer the following question:
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?
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?
Answer the following question:
When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
To which part of the electromagnetic spectrum does a wave of frequency $5 \times 10^{11} Hz$ belong?
View full solution →To which part of the electromagnetic spectrum does a wave of frequency $3 \times 10^{13} Hz$ belong?
View full solution →Answer the following question:
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?
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?
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.
View full solution →Answer the following question: 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?
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
The $6563 \mathring A \ \text{H}\alpha$ line emitted by hydrogen in a star is found to be redshifted by $15 \mathring A .$ Estimate the speed with which the star is receding from the Earth.
View full solution →Light of wavelength $5000 \mathring A $ falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
View full solution →A beam of light consisting of two wavelengths, $650 \ nm$ and $520 \ nm$, is used to obtain interference fringes in a Young’s double-slit experiment.
View full solution →- Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650 \ nm.$
- What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Answer the following question:
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.
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.
What is the shape of the wavefront in each of the following cases:
- Light diverging from a point source.
- Light emerging out of a convex lens when a point source is placed at its focus.
- The portion of the wavefront of light from a distant star intercepted by the Earth.
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?
For a single slit of width "a", the first minimum of the interference pattem of a monochromatic light of wavelength$\lambda$. Occurs at an angle of$\frac{\lambda}{\text{a}}$. At the same angle of$\frac{\lambda}{\text{a}},$ we get a maximum for two narrow slits separated by a distance "a". Explain.
View full solution →Interference is based on the superposition principle. According to this principle, at a particular point in the medium, the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.
If two sodium lamps illuminate two pinholes $S_1$ and $S_2,$ the intensities will add up and no interference fringes will be observed on the screen.
Here the source undergoes abrupt phase change in times of the order of $10^{-10}$ seconds.
View full solution →If two sodium lamps illuminate two pinholes $S_1$ and $S_2,$ the intensities will add up and no interference fringes will be observed on the screen.
Here the source undergoes abrupt phase change in times of the order of $10^{-10}$ seconds.
- Two coherent sources of intensity $\text{10 }\frac{\text{W}}{\text{m}^2}$ and $\text{25 }\frac{\text{W}}{\text{m}^2}$ interfere to form fringes. Find the ratio of maximum intensity to minimum intensity.
- $\text{y}_1=\text{a}\sin\Big[\omega\text{t}+\frac{\pi}{3}\Big]$ and $\text{y}_2=\text{a}\sin\omega\text{t}$ is:
- $\text{a}$
- $\sqrt2\text{a}$
- $\text{2a}$
- $\sqrt3\text{a}$
- The resultant amplitude of a vibrating particle by the superposition of the two waves.
- Infinite
- Five
- Three
- Zero
- The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double $-$ slit experiment, is:
- $2D$
- $4D$
- $\frac{\text{D}}{2}$
- $\frac{\text{D}}{4}$
- ln a Young's double $-$ slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen $-$ to $-$ slit distance $D$ must be changed to:
- Soap bubble.
- Excessively thin film.
- A thick film.
- Wedge shaped film.
- Which of the following does not show interference?
- $15.54$
- $16.78$
- $19.72$
- $18.39$
TV signals broadcast by Delhi studio cannot be directly received at Patna which is about 1000km away. But the same signal goes some 36000km away to a satellite, gets reflected and is then received at Patna. Explain.
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?
Huygen's principle is the basis of wave theory of light. Each point on a wavefront acts as a fresh source of new disturbance, called secondary waves or wavelets. The secondary wavelets spread out in all directions with the speed light in the given medium.
An initially parallel cylindrical beam travels in a medium of refractive index $\mu(\text{I})=\mu_0+\mu_2\text{I}$, where $\mu_0$ and $\mu_2$ are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
View full solution →An initially parallel cylindrical beam travels in a medium of refractive index $\mu(\text{I})=\mu_0+\mu_2\text{I}$, where $\mu_0$ and $\mu_2$ are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
- The initial shape of the wavefront of the beam is:
- Planar.
- Convex.
- Concave.
- Convex near the axis and concave near the periphery.
- According to Huygens Principle, the surface of constant phase is:
- Called an optical ray.
- Called a wave.
- Called a wavefront.
- Always linear in shape.
- As the beam enters the medium, it will:
- Travel as a cylindrical beam.
- Diverge.
- Converge.
- Diverge near the axis and converge near the periphery.
- Two plane wavefronts oflight, one incident on a thin convex lens and another on the refracting face of a thin prism. After refraction at them, the emerging wavefronts respectively become.
- Plane wavefront and plane wavefront.
- Plane wavefront and spherical wavefront.
- Spherical wavefront and plane wavefront.
- Spherical wavefront and spherical wavefront.
- Which of the following phenomena support the wave theory of light?
- Scattering.
- Interference.
- Diffraction.
- Velocity of light in a denser medium is less than the velocity of light in the rarer medium.
- 1, 2, 3
- 1, 2, 4
- 2, 3, 4
- 1, 3, 4
In Young’s double $-$ slit experiment using monochromatic light of wavelength $\lambda ,$ the intensity of light at a point on the screen where path difference is $\lambda ,$ is $K$ units. What is the intensity of light at a point where path difference is $\lambda /3$?
View full solution →Monochromatic light of wavelength $589 \ nm$ is incident from air on a water surface. What are the wavelength, frequency and speed of $(a)$ reflected, and $(b)$ refracted light? Refractive index of water is $1.33.$
View full solution →Let us list some of the factors, which could possibly influence the speed of wave propagation:
- Nature of the source.
- Direction of propagation.
- Motion of the source and/or observer.
- Wavelength.
- Intensity of the wave.
- The speed of light in vacuum,
- The speed of light in a medium (say, glass or water), depend?
Consider the arrangement shown in figure (17-E4). The distance D is large compared to the separation d between the slits.
- Find the minimum value of d so that there is a dark fringe at 0.
- Suppose d has this value. Find the distance x at which the next bright fringe is formed.
- Find the fringe-width.
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 $I_0$ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.
'Darkness can arise when light is mixed with light.' This phenomenon called ____________
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