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Physics M.C.Q [1M]

Light Waves · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 27 questions.

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M.C.Q [1M]

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Question 11 Mark
The equation of a light wave is written as $\text{y}=\text{A}\ \sin(\kappa\text{x}-\omega\text{t}).$ Here, y represents:
  1. Displacement of either particles.
  2. Pressure in the medium.
  3. Density of the medium.
  4. Electric field.
Answer
  1. Electric field.

Explanation:

Light consists of mutually perpendicular electric and magnetic fields. So, the equation of a light wave is represented by its field vector.

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Question 21 Mark
The wavefronts of a light wave travelling in vacuum are given by x + y + z = c. The angle made by the direction of propagation of light with the X-axis is:

  1. $0^\circ$

  2. $45^\circ$

  3. $90^\circ$

  4. $\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

Answer
  1. $\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

Explanation:

On writing the given equation in the plane equation form lx + my + nz = p,

Where l2 + m2 + n2 and p > 0, we get:

$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{\text{c}}{\sqrt{3}}$

If $\theta$ is the angle between the normal and +x axis, then

$\cos\theta=\frac{1}{\sqrt{3}}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big) $

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Question 31 Mark
Three observers A, B and C measure the speed of light coming from a source to be vA, 0B and vc. The observer A moves towards the source and C moves away from the source at the same speed. The observer B stays stationary. The surrounding space is vacuum everywhere.

  1. $\text{v}_\text{A}>\text{v}_\text{B}>\text{v}_\text{C}.$

  2. $\text{v}_\text{A}<\text{v}_\text{B}<\text{v}_\text{C}$

  3. $\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}$

  4. $\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$

Answer

  1. $\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}$

  2. $\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$

Explanation:

Since the speed of light is a universal constant, $\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}=3\times10^8\text{m/s.}$

$\text{v}_\text{B}=\frac{1}{2}(\text{u}_\text{A}+\text{u}_\text{C})$ This expression also implies that vA = vB = vC.

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Question 41 Mark
If the source of light used in a Young's double slit experiment is changed from red to violet:
  1. The fringes will become brighter.
  2. Consecutive fringes will come closer.
  3. The intensity of minima will increase.
  4. The central bright fringe will become a dark fringe.
Answer
  1. Consecutive fringes will come closer.

Explanation:

Fringe width, $\beta=\frac{\lambda\text{D}}{\text{d}}$

Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

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Question 51 Mark
A light wave can travel:
  1. In vacuum.
  2. In vacuum only.
  3. In a material medium.
  4. In a material medium only
Answer
  1. In vacuum.
  1. In a material medium.

Explanation:

Light is an electromagnetic wave that can travel through vacuum or any optical medium.

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Question 61 Mark
A thin transparent sheet is placed in front of a Young's double slit. The fringe-width will:
  1. Increase.
  2. Decrease.
  3. Remain same.
  4. Become nonuniform.
Answer
  1. Remain same.

Explanation:

On the introduction of a transparent sheet in front of one of the slits, the fringe pattern will shift slightly but the width will remain the same.

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Question 71 Mark
A Young's double slit experiment is performed with white light:
  1. The central fringe will be white.
  2. There will not be a completely dark fringe.
  3. The fringe next to the central will be red.
  4. The fringe next to the central will be violet.
Answer

  1. The central fringe will be white.

  2. There will not be a completely dark fringe.

  1. The fringe next to the central will be violet.

Explanation:

The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.

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Question 81 Mark
Light is:
  1. Wave phenomenon.
  2. Particle phenomenon
  3. Both particle and wave phenomenon.
Answer
  1. Both particle and wave phenomenon.

Explanation:

Light shows photoelectric effect and Compton effect, which depicts its particle nature. It also shows interference and diffraction, which depicts the wave nature of light.

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Question 91 Mark
Which of the following properties show that light is a transverse wave?
  1. Reflection.
  2. Interference.
  3. Diffraction.
  4. Polarization.
Answer
  1. Polarization.

Explanation:

Reflection, interference and diffraction are the phenomena shown by both transverse waves and longitudinal waves. Polarization is the phenomenon shown only by transverse waves.

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Question 101 Mark
When light is refracted, which of the following does not change?
  1. Wavelepgth.
  2. Frequency.
  3. Velocity.
  4. Amplitude.
Answer
  1. Frequency.

Explanation:

Frequency of a light wave doesnt change on changing the medium of propagation of light.

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Question 111 Mark
When a drop of oil is spread on a water surface, it displays beautiful colours in daylight because of:
  1. Disperson of light.
  2. Reflection of light.
  3. Polarization of light.
  4. Interference of light.
Answer
  1. Interference of light.

Explanation:

Interference effect is produced by a thin film (coating of a thin layer of a translucent material on a medium of different refractive index which allows light to pass through it). ln the present case, oil floating on water forms a thin film on the surface of water, leading to the display of beautiful colours in daylight because of the interference of sunlight.

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Question 121 Mark
Which of the following sources gives best monochromatic light?
  1. A candle.
  2. A bulb.
  3. A mercury tube.
  4. A laser.
Answer
  1. A laser.

Explanation:

Among the given sources, laser is the best coherent source providing monochromatic light with constant phase difference.

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Question 131 Mark
The amplitude modulated (AM) radio wave bends appreciably round the corners of a 1m × 1m board but the frequency modulated (FM) wave only negligibly bends. If the average wavelengths of AM and FM waves are $\lambda_\text{a}$ and $\lambda_\text{f}:$

  1. $\lambda_\text{a}>\lambda_\text{f}$

  2. $\lambda_\text{a}=\lambda_\text{f}$

  3. $\lambda_\text{a}<\lambda_\text{f}$

  4. We don't have sufficient information to decide about the relation of $\lambda_\text{a}$ and $\lambda_\text{f}$

Answer
  1. $\lambda_\text{a}>\lambda_\text{f}$

Explanation:

An electromagnetic wave bends round the corners of an obstacle if the size of the obstacle is comparable to the wavelength of the wave. An AM wave has less frequency than an FM wave, So, an AM wave has a higher wavelength than an FM wave and it bends round the comers of a 1m × 1m board.

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Question 141 Mark
Two sources are called coherent if they produce waves:
  1. Of equal wavelength.
  2. Of equal velocity.
  3. Having same shape of wavefront.
  4. Having a constant phase difference.
Answer
  1. Having a constant phase difference.

Explanation:

For light waves emitted by two sources of light to remain coherent, the initial phase difference between waves should remain constant in time. If the phase difference changes continuously or randomly with time, then the sources are incoherent.

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Question 151 Mark
Light waves travel in vacuum along the X-axis. Which of the following may represent the wavefronts?
  1. x = c.
  2. y = c.
  3. z = c.
  4. x + y + z = c.
Answer
  1. x = c.

Explanation:

​The wave is travelling along the X-axis. So, it'll have planar wavefront perpendicular to the X-axis.

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Question 161 Mark
Four light waves are represented by

  1. $\text{y}=\text{a}_1\sin\omega\text{t}$

  2. $\text{y}=\text{a}_2\sin(\omega\text{t}+\in)$

  3. $\text{y}=\text{a}_1\sin2\omega\text{t}$

  4. $\text{y}=\text{a}_2\sin2(\omega\text{t}+\in)$

Interference fringes may be observed due to superposition of:

  1. (i) and (ii)
  2. (i) and (iii)
  3. (ii) and (iv)
  4. (iii) and (iv)
Answer
  1. (i) and (ii)
  1. (iii) and (iv)

Explanation:

The waves are travelling with the same frequencies and varying by constant phase difference. 

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Question 171 Mark
The wavefronts of light coming from a distant source of unknown shape are nearly:
  1. Plane.
  2. Elliptical.
  3. Cylindrical.
  4. Spherical.
Answer
  1. Plane.

Explanation:

Wave travelling from a distant source always has plane wavefront.

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Question 181 Mark
When light propagates in vacuum there is an electric field and a magnetic field. These fields:
  1. Are constant in time.
  2. Have zero average value.
  3. Are perpendicular to the direction of propagation of light.
  4. Are mutually perpendicular.
Answer
  1. Have zero average value.
  2. Are perpendicular to the direction of propagation of light.
  3. Are mutually perpendicular.

Explanation:

Light is an electromagnetic wave that propagates through its electric and magnetic field vectors, which are mutually perpendicular to each other, as well as to the direction of propagation of light. The average value of both the fields is zero.

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Question 191 Mark
Huygens' principle of secondary wavelets may be used to:
  1. Find the velocity of light in vacuum.
  2. Explain the particle behaviour of light.
  3. Find the new position of a wavefront.
  4. Explain Snell's law.
Answer
  1. Find the new position of a wavefront.
  2. Explain Snell's law.

Explanation:

Huygen's wave theory explains the origin of points for the new wavefront proceeding successively. It also explains the variation in speed of light on moving from one medium to another, i.e. it proves Snell's Law.

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Question 201 Mark
Which of the following properties of light conclusively support wave theory of light?
  1. Light obeys laws of reflection.
  2. Speed of light in water is smaller than the speed in vacuum.
  3. Light shows interference.
  4. Light shows photoelectric effect.
Answer
  1. Speed of light in water is smaller than the speed in vacuum.
  2. Light shows interference.

Explanation:

Snell's Law, which states that the speed of light reduces on moving from a rarer to a denser medium, can be concluded from Huygens' wave theory and interference of light waves is based on the wave properties of light.

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Question 211 Mark
Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. The intensities of the sources are in the ratio:
  1. 25 : 1
  2. 5 : 1
  3. 9 : 4
  4. 625 : 1
Answer
  1. 9 : 4

Explanation:

Ratio of maximum intensity and minimum intensity is given by

$\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(\sqrt{\text{I}_1}+\sqrt{\text{I}_2})^2}{(\sqrt{\text{I}_1}-\sqrt{\text{I}_2})^2}=\frac{25}{1}$

$\Rightarrow\sqrt{\text{I}_1}=3 \ \text{and}\ \sqrt{\text{I}_2}=2$

$\Rightarrow\text{I}_1=9\ \text{and}\ \text{I}_2=4$

Then,

$\frac{\text{I}_1}{\text{I}_2}=\frac{9}{4}$

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Question 221 Mark
If Young's double slit experiment is performed in water:
  1. The fringe width will decrease.
  2. The fringe width will increase.
  3. The fringe width will remain unchanged.
  4. There will be no fringe.
Answer
  1. The fringe width will decrease.

Explanation:

As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,

Here,

$\lambda_\text{M}=\frac{\lambda}{\eta}$

$\lambda_\text{M}$ = wavelength in medium

$\lambda$ = wavelength in vacuum

$\eta$ = refractive index of medium

Hence, fringe width decreases when Young's double slit experiment is performed under water.

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Question 231 Mark
When light is refracted into a medium:
  1. Its wavelength and frequency both increase
  2. Its wavelength increases but frequency remains unchanged
  3. Its wavelength decreases but frequency remains unchanged.
  4. Its wavelength and frequency both decrease.
Answer
  1. Its wavelength decreases but frequency remains unchanged.

Explanation:

Frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.

Decrease in the wavelength of light entering a medium of refractive index $\mu$, is given by,

$\lambda_\text{M}=\frac{\lambda}{\mu},$

Where $\lambda_\text{M}$ = wavelength in medium

$\lambda$ = wavelength in vacuum

$\mu$ = refractive index

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Question 241 Mark
The inverse square law of intensity $\Big(\text{i.e., the intensity}\propto\frac{1}{\text{r}^2}\Big)$ is valid for a:
  1. Point source.
  2. Line source.
  3. Plane source.
  4. Cylindrical source.
Answer
  1. Point source.

Explanation:

Intensity of a point source obeys the inverse square law.

Intensity of light at distance r from the point source is given by

$\text{I}=\frac{\text{S}}{(4\pi\text{r}^2)}$

Where S is the source strength.

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Question 251 Mark
The speed of light depends:
  1. On elasticity of the medium only.
  2. On inertia of the medium only.
  3. On elasticity as well as inertia.
  4. Neither on elasticity nor on inertia.
Answer
  1. Neither on elasticity nor on inertia.

Explanation:

The speed of light in any medium depends on the refractive index of that medium, which is an intensive property. Hence, speed of light is not affected by the elasticity and inertia of the medium.

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Question 261 Mark
The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be:

  1. $\text{I}_0$

  2. $\frac{\text{I}_0}{4}$

  3. $\frac{\text{I}_0}{2}$

  4. $4\text{I}_0$

Answer
  1. $\frac{\text{I}_0}{4}$

Explanation:

Total intensity coming from the source is I0 which is present at the central maxima. In case of two slits, the intensity is getting distributed between the two slits and for a single slit, the amplitude of light coming from the slit is reduced to half which leads to $\frac{1}{4}\text{th}$ of intensity.

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Question 271 Mark
Suppose the medium in the previous question is water. Select the correct option(s) from the list given in that question.
  1. vA > vB > vC
  2. vA < vB < vC
  3. vA = vB = vC
  4. $\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$
Answer

  1.  $\text{v}_\text{A}>\text{v}_\text{B}>\text{v}_\text{C}$

  2. $\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$

Explanation:

Since the speed of light is a universal constant,

$\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}=3\times 10^8\text{m/s}$

$\text{v}_\text{B}=\frac{1}{2}(\text{u}_\text{A}+\text{u}_\text{C})$ This expression also implies that vA = vB = vC 

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