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Physics M.C.Q [1M]

Wave Optics · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 177 questions.

Questions · Page 2 of 4

M.C.Q [1M]

MCQ 511 Mark
Consider sunlight incident on a slit of width 104A. The image seen through the slit shall.
  • A
    Be a fine sharp slit white in colour at the center.
  • B
    A bright slit white at the center diffusing to zero intensities at the edges.
  • C
    A bright slit white at the center diffusing to regions of different colours.
  • D
    Only be a diffused slit white in colour.
Answer
  1. Be a fine sharp slit white in colour at the center.

Solution:

Key concept:

Diffraction of Light is the phenomenon of bending of light around the comers of an obstacle/aperture of the size of the wavelength of light.

Size of the slit is very large compared to wavelength

Size of the slit is comparable to wavelenght

In figure (A), no diffraction phenomenon is observed as the size of slit is weary large compared to wavelength. But in figure(B), there will be diffraction of light as size of slit is compared to the wavelength of light incident.

Here in the question it is given, width of the slit

$\text{b}=10^4\mathring{\text{A}}=10^4\times10^{-10}\text{m}$

$=10^{-6}\text{m}=1\text{pm}$

Wavelength of (visible) sunlight varies from $4000\mathring{\text{A}}\text{ to }8000\mathring{\text{A}}$.

Hence the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear, i.e., mixing of colours form white patch at the centre.

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MCQ 521 Mark
When a drop of oil is spread on a water surface, it displays beautiful colours in daylight because of:
  • A
    Disperson of light.
  • B
    Reflection of light.
  • C
    Polarization of light.
  • D
    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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MCQ 531 Mark
Light waves travel in vaccum along the y-axis. Then the wave front is:
  • A
    y = constant
  • B
    x = constant
  • C
    z = constant
  • D
    x+y+z = constant
Answer
  1. y = constant

Explanation:

waves front are plane perpendicular to the direction of rays.

so, as light is traveling along y - axis, plane perpendicular to y - axis is the x-z plane with any constant value of y.

so, y = constant is the wave front plane

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MCQ 541 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:
  • A
    $0^\circ$
  • B
    $45^\circ$
  • C
    $90^\circ$
  • D
    $\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
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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MCQ 551 Mark
The phenomenon of rotation of plane polarized light is called:
  • A
    Kerr effect
  • B
    Double refraction
  • C
    Optical activity
  • D
    Dichroism
Answer
  1. Optical activity

Explanation:

The phenomenon of rotation of plane polarized light is called optical activity.

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MCQ 561 Mark
In which of the following the final image is erect? 
  • A
    Compound microscope
  • B
    Astronomical telescope
  • C
    Simple microscope
  • D
    All of the above
Answer
  1. Simple microscope

Explanation:

The image formed by the Compound microscope and Astronomical telescope is inverted,but in case of Simple microscope it form erect image.

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MCQ 571 Mark
A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plae as shown in figure. The observed interference fringes from this combination shall be:

  • A
    Straight
  • B
    Circular
  • C
    Equally spaced
  • D
    Having fringe spacing which increases as we go outward.
Answer
  1. Straight

Explanation:

The locus of the equal path difference consists in lines going parallel to the axis of cylinder. 

Therefore, interference fringes will be straight.

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MCQ 581 Mark
Two identical lights sources S1​ and S2​ emit the light of same wavelength $\lambda$. These light rays will exhibit interference if:
  • A
    Their phase difference remain constant.
  • B
    Their phase difference is distributed randomly.
  • C
    Their light intensities remain constant.
  • D
    Their light intensities change continuously.
Answer
  1. Their phase difference remain constant.

Explanation:

For interference to take place the light sources need to be either in phase or have a constant phase difference. In case the phase difference keeps changing the interference pattern will keep on changing, as a result of interference pattern will be observed.

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MCQ 591 Mark
When a drop of oil is spread on a water surface, it displays beautiful colors in daylight because of:
  • A
    Dispersion of light
  • B
    Reflection of light
  • C
    Polarization of light
  • D
    Interference of light
Answer
  1. Interference of light

Explanation:

When a drop of oil is spread on a water surface, it displays beautiful colors in daylight because the oil film is only a few nano-meters thick. Some of the light is reflected off the top surface and some the bottom surface. Because the thickness of the oil film is about the same as the wavelength of the light the two reflected rays interfere with each other.

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MCQ 601 Mark
If the apertature of a telescope is decreased the resolving power will:
  • A
    Increases
  • B
    Decreases
  • C
    Remain same
  • D
    Zero
Answer
  1. Decreases

Explanation:

Resolving power of a telescope $=\frac{\text{a}}{1.22\lambda}$

where a is the aperture of the telescope.

Thus resolving power∝ aperture.

Hence, if the aperture of telescope decreases, the resolving power decreases.

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MCQ 611 Mark
In a Young's double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case:
  • A
    There shall be alternate interference patterns of red and blue.
  • B
    There shall be an interference pattern for red distinct from that for blue.
  • C
    There shall be no interference fringes.
  • D
    There shall be an interference pattern for red mixing with one for blue.
Answer
  1. There shall be no interference fringes.

Solution:

We know that, for the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength.

In a Young's double-slit experiment, if one of the holes is covered by a red filter and another by a blue filter, then only red and blue lights are present due to alteration. In Young's double-slit experiment, a monochromatic light is used for the formation of fringes on the screen.

Therefore, there shall be no interference fringes.

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MCQ 621 Mark
Two source S1 and S2 of intensity I1 and I2 are placed in front of a screen [Fig. (a)]. The patteren of intensity distribution seen in the central portion is given by Fig. (b).

In this case which of the following statements are true.
  • A
    S1 and S2 have the same intensities.
  • B
    S1 and S2 have a constant phase difference.
  • C
    S1 and S2 have the same phase.
  • D
    S1 and S2 have the same wavelength.
Answer
  1. S1 and S2 have the same intensities.
  2. S1 and S2 have a constant phase difference.
  1. S1 and S2 have the same wavelength.

Solution:

Key concept:

  1. For getting the sustained interference the initial phase difference between the interfering waves must remain constant, i.e., sources should be coherent.

For two coherent sources, the resultant intensity is given bt $\text{I}=\text{I}_1+\text{I}_2+2\sqrt{\text{I}_1\text{I}_2}\cos\phi$.

  1. Resultant intensity at the point of observation will be maximum.

$\text{I}_{\text{max}}=\text{I}_1+\text{I}_2+2\sqrt{\text{I}_1\text{I}_2}$

$\text{I}_\text{max}=(\sqrt{\text{I}_1}+\sqrt{\text{I}_2})^2$

  1. Resultant intensity at the point of observation will be minimum.

$\text{I}_{\text{max}}=\text{I}_1+\text{I}_2+2\sqrt{\text{I}_1\text{I}_2}$

$\text{I}_\text{max}=(\sqrt{\text{I}_1}+\sqrt{\text{I}_2})^2$

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MCQ 631 Mark
Wave front means:
  • A
    All particles in it have same phase.
  • B
    Few particles are in same phase, rest are in opposite phase.
  • C
    All particles have opposite phase of vibrations.
  • D
    All particles have random vibrations.
Answer
  1. All particles in it have same phase.

Explanation:

It is the imaginary surface representing corresponding points pf a wave that vibrate in unison. When identical waves having a common origin travel through a homogeneous medium, the corresponding crests and troughs at any instant phase.

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MCQ 641 Mark
In Young's interference experiment, if the slits are of unequal width, then:
  • A
    No fringes will be formed
  • B
    The positions of minimum intensity will not be completely dark.
  • C
    Bright fringe as displaced from the original central position.
  • D
    Distance between two consecutive dark fringes will not be equal to the distance between two consecutive right fringes.
Answer
  1. The positions of minimum intensity will not be completely dark.

Explanation:

Unequal width of slits will cause unequal intensity of lights entering from both slits.

As a result, during interference complete cancelling of light intensity will not take place at regions of otherwise dark fringe.

As the value of $\beta$ does not depend on intensity of light, there will be no shifting of fringes as well as no change in fringe width.

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MCQ 651 Mark
Light waves can be polarised because they:
  • A
    Have high frequencies
  • B
    Have short wavelength
  • C
    Are transverse
  • D
    Can be reflected
Answer
  1. Are transverse

    Explanation:

    Polarisation of light waves is possible only because they can oscillate in more than one orientation i.e., they are transverse in nature. It has no dependence on its wavelength and frequencies.

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MCQ 661 Mark
A person shines a coherent light of a bulb through an object, A, which produces a pattern of concentric rings on a screen, B. A is most likely:

  • A
    A polarization filter
  • B
    A single-slit
  • C
    A prism
  • D
    A sheet with a pinhole
Answer
  1. A sheet with a pinhole

Explanation:

A polarization filter will limit the light vectors in a single plane i.e. polarization.

A prism will split the light i.e. dispersion .

A slit system will make a diffraction pattern on screen B, but in this pattern, fringes are straight.

Due to its circular size, a pinhole will produce diffraction pattern on screen B in the forms of concentric rings.

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MCQ 671 Mark
A white light is passed through the two narrow slits and produced a pattern of alternating bright and dark lines on the screen as shown above. What will effects on the central bright band , if the distance between screen and slits are increased?

  • A
    Become wider
  • B
    Become narrower
  • C
    Remain unchanged
  • D
    Disappear completely
Answer
  1. Become wider

Explanation:

The setup shown is that of Young's Double Slit Experiment.

The fringe width in the experiment is given as $\beta=\frac{\text{D}\lambda}{\text{d}}​$

where d is the distance between the slits, D is the distance between screen and slits.

Hence the central fringe also widens when distance between screen and slits increases.

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MCQ 681 Mark
The magnitude of magnifying power of an astronomical telescope is 5, the focal power of its eyepiece is 10 diopters. The focal power of its objective (in diopters) is:
  • A
    1
  • B
    2
  • C
    3
  • D
    4
Answer
  1. 2

Explanation:

M = 5

Pe = 10

$\frac{1}{\text{f}_\text{e}}=10$

fe​ = $\frac{1}{10}$m = 10cm

M = $\frac{\text{f}_\text{0}}{\text{f}_\text{e}}$​​ =5

f0 ​= 50cm

P = 2D

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MCQ 691 Mark
Which of the following is correct for light diverging from a point source?
  • A
    The intensity decreases in proportion for the distance squared.
  • B
    The wavefront is parabolic.
  • C
    The intensity at the wavelength does not depend on the distance.
  • D
    None of these.
Answer
  1. The intensity decreases in proportion for the distance squared.

Explanation:

A point source produces a spherical wavefront that moves in all the directions from the point.

Whereas the intensity at the wavelength is dependent on the distance and it follows the inverse square law. So, the intensity decreases with an increase in the distance from the source.

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MCQ 701 Mark
In YDSE, the slit widths are in the ratio of 1:9. The ratio of intensity of the maxima to that of the minima is:
  • A
    81:1
  • B
    9:1
  • C
    4:1
  • D
    3:1
Answer
  1. 4:1

Explanation:

Intensity is proportional to the area of the slit.

As slit widths are in the ratio of 1:9

The areas are also in the ratio 1:9

Thus Intensities are in the ratio 1:9

amplitudes are square root of Intensities

Thus amplitudes are in ratio 1:3

Let amplitudes be x and 3x

At maxima the amplitudes get added up x + 3x = 4x

At minima they become x − 3x = −2x

Intensity of maxima to minima is $\frac{16\text{x}^2}{4\text{x}^2}=\frac{4}{1}$

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MCQ 711 Mark
If a star emitting yellow light is accelerated towards earth, then to an observer on earth it will appear?
  • A
    Becoming orange
  • B
    Shining yellow
  • C
    Gradually changing to blue
  • D
    Gradually changing to red
Answer
  1. Gradually changing to blue

Explanation:

As the star coming closer to the earth, the frequency of the light increase and the wavelength decrease due to doppler effect. So the colour should gradually change to blue.

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MCQ 721 Mark
Making a light wave vibrate in only one plane is known as:
  • A
    Refraction.
  • B
    Interference
  • C
    Diffraction
  • D
    Polarization.
Answer
  1. Polarization.

Explanation:

Light in the form of a plane wave in space is said to be linearly polarized. Light is a transverse electromagnetic wave, but natural light is generally unpolarized, all planes of propagation being equally probable.

The process of making the light wave vibrated in a single plane is known as polarization.

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MCQ 731 Mark
The interfering fringes formed by a thin oil film on water are seen in yellow light of sodium lamp. We find the fringes:
  • A
    Coloured
  • B
    Black and white
  • C
    Yellow and black
  • D
    Coloured white yellow
Answer
  1. Yellow and black

Explanation:

When the yellow light of sodium lamp interferes constructively we get yellow bright band, and when they interfere destructively we get black bark band.

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MCQ 741 Mark
Who amongst the following used corpuscular theory to explain the nature of light?
  • A
    Max Planck
  • B
    Newton
  • C
    Young
  • D
    Einstein
Answer
  1. Newton

Explanation:

Newton was the first to use corpuscular theory to explain the nature of light. He was able to successfully reflection and refraction using this theory.

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MCQ 751 Mark
In Young's double slit experiment, the interference pattern obtained with white light will be:
  • A
    The central fringe bright and alternate bright and dark fringes.
  • B
    The central fringe achromatic and coloured fringes for small path difference.
  • C
    The central fringe dark
  • D
    The central fringe coloured
Answer
  1. The central fringe achromatic and coloured fringes for small path difference.

Explanation:

If white light is used a white centre fringe is observed, but all the other fringes have coloured edges, the blue edge being nearer the centre. Eventually the fringes overlap and a uniform white light is produced.

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MCQ 761 Mark
Which one of the following is more monocromatic?
  • A
    Laser beam
  • B
    White light
  • C
    Sodium light
  • D
    Mercury light
Answer
  1. Laser beam

Explanation:

Monochromatic light is light made up of one single pure frequency.white light, which is light that contains all frequencies, That means the sum of red, yellow, green, blue and violet. Some light sources send out monochromatic light such as lasers beam.

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MCQ 771 Mark
The diameter of objective of a telescope is 1m. Its resolving limit for the light of wavelength 4538$\mathring{\text{A}}$, will be:
  • A
    5.54×10−7rad
  • B
    2.54×10−4rad
  • C
    6.54×10−7rad
  • D
    None of the above
Answer
  1. 5.54×10−7rad

Explanation:

Resolving limit

$\text{d}\theta=\frac{1.22\lambda}{\text{a}}$

$=\frac{1.22\times4538\times10^{-10}}{1}$

= 5.54×10−7rad.

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MCQ 781 Mark
A man wishing to get a picture of a Zebra photographed a white donkey after fitting a glass with black streaks onto the objective of his camera.
  • A
    The image will look like a white donkey on the photograph.
  • B
    The image will look like a Zebra on the photograph.
  • C
    The image will be more intense compared to the case in which no such glass is used.
  • D
    The image will be less intense compared to the case in which no such glass is used.
Answer
  1. The image will look like a Zebra on the photograph.

Explanation:

The rays coming from the body of the white donkey will interfere with the black streaks on the glass and the final image of a zebra will be produced.

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MCQ 791 Mark
Doppler's effect is sound in addition of relative velocity between source and observer, also depends while source and observer or both are moving. Doppler's effect in light depend only on the relative velocity of source and observer. The reason of this is.
  • A
    Einstein mass-energy relation.
  • B
    Einstein theory of relativity
  • C
    Photoelectric effect
  • D
    None of the above
Answer
  1. None of the above

Explanation:

Doppler effect refers to the change in wave frequency during the relative motion between a wave source and it's observer by considering that the Velocity of the observer with respect to the source velocity is negligible.

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MCQ 801 Mark
Find out correct option which describe the Red shift of distant galaxies.
  • A
    Expansion of the universe
  • B
    The Uncertainty Principle
  • C
    Black holes
  • D
    Superconductivity
Answer
  1. Expansion of the universe

Explanation:

Red shift is the increase in wavelength of light observed due to Doppler Effect of light, observed when objects recede from each other.

The red shift as observed in distant galaxies suggests that the galaxies are moving away from us and from each other. This tells that the universe is expanding.

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MCQ 811 Mark
The wave front due to a source situated at infinity is:
  • A
    Spherical
  • B
    Cylindrical
  • C
    Planar
  • D
    None of these
Answer
  1. Planar

Explanation:

when you considered it a large distance and measuring justice Mall section of it then it can be considered to be plane wavefront source at Infinity example the one coming from sun to earth surface is considered to be plain VU friend from light diverging from a point source will be spherical.

So, the wave front due to a source situated at infinity is planar.

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MCQ 821 Mark
Which of the following statements about the behaviour of light is not correct?
  • A
    Interference patterns are evident for light behaving as rays.
  • B
    Ray properties of light are useful for understanding how images are formed by optical devices such as eyes.
  • C
    Wave properties are important for observing the behaviour of light at a fine scale.
  • D
    Both wave and particle theories of light can be related to the colour sensations produced by light.
Answer
  1. Interference patterns are evident for light behaving as rays.

Explanation:

Interference patterns are explained using wave nature of light. You can learn more from youtube video "Interference of light".

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MCQ 831 Mark
Consider a light beam incident from air to a glass slab at Brewster's angle as shown in Fig. A polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.

  • A
    For a particular orientation there shall be darkness as observed through the polaoid.
  • B
    The intensity of light as seen through the polaroid shall be independent of the rotation.
  • C
    The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.
  • D
    The intensity of light as seen through the polaroid shall go through a minimum for four orientations of the polaroid.
Answer
  1. The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.

Solution:

Hint: If a light beam incidents at Brewster's angle, then the transmitted beam is always unpolarised and reflected beam is always polarised.

In the given diagram, the light beam incident from air to the glass slab at Brewster's angle (ip). Therefore, the incident ray represented by dot (.), is unpolarised and the reflected light represented by arrow, is plane polarized.

Since, the emergent ray is unpolarised.

Hence, the intensity cannot be zero when passes through polaroid.

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MCQ 841 Mark
In Young's double slit experiment the intensity of the maxima is I. If the width of each slit is doubled, the intensity of the maxima will be:
  • A
    $\frac{\text{I}}{2}$
  • B
    2I
  • C
    4I
  • D
    I
Answer
  1. 2I

Explanation:

I = Imax​ = 4I0​

Now, I0​ is increased to 2I0​

So, Imax ​= 4(2I0​) = 8I0​ = 2I

So, maximum intensity is 2I

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MCQ 851 Mark
For the propagation of light wave, medium is required. This is according to:
  • A
    Maxwell's theory
  • B
    Huygen's theory
  • C
    Planck's theory
  • D
    Newton's theory
Answer
  1. Huygen's theory

Explanation:

Huygens suggested that light may be a wave phenomenon produced by mechanical vibrations of an all pervading hypothetical homogenous medium called eather just like those in solids and liguid .This medium was supposed to be mass less with extremely high elasticity and very low density.

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MCQ 861 Mark
The phenomenon by which stars recedes from each other is explained by:
  • A
    Black hole theory
  • B
    Neutron star theory
  • C
    White dwarf
  • D
    Red shift
Answer
  1. Red shift

Explanation:

When an object moves away from us, its light waves are stretched into lower frequencies or longer wavelengths, and we say that the light is redshifted. It also explain the expanding nature of universe. 

Doppler effect in light explains the phenomenon of receding of stars and approaching of star by red shift and blue shift respectively.

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MCQ 871 Mark
Consider the diffraction patern for a small pinhole. As the size of the hole is increased:
  • A
    The size decreases.
  • B
    The intensity increases.
  • C
    The size increases.
  • D
    The intensity decreases.
Answer
  1. The size decreases.
  2. The intensity increases.

Solution:

Key concept: The "shadow" of hole of diameter d is spread out over an angle $\Delta\theta=1.22\frac{\lambda}{\text{D}}\Rightarrow\ \Delta\theta\propto\frac{1}{\Delta}$

The central bright disc is known as Airy's disc.

As the size of the hole is increased, AO will decrease and size of Airy's disc will decrease.

As the size of the hole is increased, the width of the central maximum of the diffraction pattern of hole decreases. Since the same amount of light is now distributed over a small area, as intensity ∞ 1/area, the area is decreasing so area intensity increases.

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MCQ 881 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.
  • A
    $\text{v}_\text{A}>\text{v}_\text{B}>\text{v}_\text{C}.$
  • B
    $\text{v}_\text{A}<\text{v}_\text{B}<\text{v}_\text{C}$
  • C
    $\text{v}_\text{A}=\text{v}_\text{B}=\text{v}_\text{C}$
  • D
    $\text{v}_\text{B}=\frac{1}{2}(\text{v}_\text{A}+\text{v}_\text{C})$
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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MCQ 891 Mark
When exposed to sunlight, thin films of oil on water often exhibit brilliant colours due to the phenomenon of:
  • A
    The dispersion
  • B
    The interference
  • C
    The diffraction
  • D
    The angular acceleration
Answer
  1. The interference

Explanation:

When exposed to sunlight, thin films of oil on water often exhibit brilliant colors due to the phenomenon of interference.

In the case of a thin oil film, a layer of oil sits atop a layer of water. The oil may have an index of refraction near 1.5 and the water has an index of 1.33.

The materials on either side of the oil film (air and water) both have refractive indices that are less than the index of the film.

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MCQ 901 Mark
Constructive and destructive interference occur in:
  • A
    Cosmic rays
  • B
    Light raus
  • C
    Sound waves
  • D
    All of these
Answer
  1. All of these

Explanation:

Interference (constructive and destructive) is one of the basic properties of a wave, therefore it will occur in all the given waves, no matter they are electromagnetic or mechanical, transverse or longitudinal.

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MCQ 911 Mark
A Young's double slit experiment is performed with white light:
  • A
    The central fringe will be white.
  • B
    There will not be a completely dark fringe.
  • C
    The fringe next to the central will be red.
  • D
    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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MCQ 921 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:
  • A
    25 : 1
  • B
    5 : 1
  • C
    9 : 4
  • D
    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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MCQ 931 Mark
Light transmitted by nicol prism is:
  • A
    Unpolarised
  • B
    Plane polarised
  • C
    Circular polarised
  • D
    Elliptically polarised
Answer
  1. Plane polarised

Explanation:

Nicol prism is a polariser in which the O-ray is eliminated by total internal reflection and the light transmitted through it, is E-ray which is completely plane polarised light.

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MCQ 941 Mark
The angular spread of central maximum, in the diffraction pattern, does not depend on ______.
  • A
    The distance between the slit and sources
  • B
    Width of slit
  • C
    Wavelength of light
  • D
    Frequency of light
Answer
  1. The distance between the slit and sources

Explanation:

Angular spread of central maxima $=(\theta)=\frac{\lambda}{\text{b}}​$

Where $\lambda$ = wavelength of light

b = width of slit

⇒ Clearly, it does depends on the distance between slit and sources.

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MCQ 951 Mark
The angular resolution of a telescope of 10cm diameter at a wavelength of 5000$\mathring{\text{A}}$ is of the order of:
  • A
    106 rad
  • B
    10−2 rad
  • C
    10−4 rad
  • D
    10−5 rad
Answer
  1. 10−5 rad

Explanation:

$\text{R}=\frac{1}{\Delta\theta}$

$=\frac{\text{a}}{1.22\lambda}$

$\frac{1}{\Delta\theta}=\frac{0.10}{1.22\times5000\times10^{-10}}$

$\Delta\theta=6.1\times10^{-6}\text{rad}$

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MCQ 961 Mark
If the source of light used in a Young's double slit experiment is changed from red to violet:
  • A
    The fringes will become brighter.
  • B
    Consecutive fringes will come closer.
  • C
    The intensity of minima will increase.
  • D
    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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MCQ 971 Mark
Huygen's concept of wavelets is useful in:
  • A
    Explaining polarisation
  • B
    Determining focal length of lenses
  • C
    Determining chromatic aberration
  • D
    Geometrical reconstruction of a wavefront
Answer
  1. Geometrical reconstruction of a wavefront

Explanation:

Huygens considered that light was propagated in longitudinal waves.

Huygen's concept explained the direction of propagation of light waves by geometrical reconstruction of wavefront.

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MCQ 981 Mark
The resolving power of a telescope depends on:
  • A
    Length of telescope
  • B
    Focal length of objective
  • C
    Diameter of the objective
  • D
    Focal length of eyepiece
Answer
  1. Diameter of the objective

Explanation:

Resolving power of telescope $\text{R}=\frac{1}{\Delta\theta}=\frac{\text{a}}{1.22\lambda}$

where, $\Delta\theta$ is angular separation between two objects.

a is the diameter of the objective.

$\lambda$ is wavelength of light.

So, clearly resolving power of a telescope depends on diameter of the objective.

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MCQ 991 Mark
The interference of light was first demonstrated experimentally by:
  • A
    Sir Isaac Newton
  • B
    Michelson
  • C
    Fraunhoffer
  • D
    Thomas Young
Answer
  1. Thomas Young

Explanation:

Thomas Young demonstrated the phenomenon of interference in water waves.

In 1801, he presented a famous paper to the Royal Society entitled "On the Theory of Light and Colours" which described various interference phenomena, and in 1803 he performed his famous double-slit experiment.

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MCQ 1001 Mark
Two lenses of focal lengths +100cm and +5cm are used to prepare an astronomical telescope. The minimum tube length will be : (final image is at ∞)
  • A
    95cm
  • B
    100cm
  • C
    105cm
  • D
    500cm
Answer
  1. 105cm

Explanation:

The length of telescope = focal length of object (−f0​) +focal length of eyepiece (fe​) = 100 + 5 = 105cm

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