Optics MCQ Quiz - Objective Question with Answer for Optics - Download Free PDF

Last updated on May 30, 2025

Latest Optics MCQ Objective Questions

Optics Question 1:

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is : 
qImage6821ae5550db2c0e03b34c1a

  1. \(-\frac{\alpha}{2}\)
  2. \(-45^\circ \)
  3. \( +45^\circ\)
  4. \(-\alpha\)

Answer (Detailed Solution Below)

Option 2 : \(-45^\circ \)

Optics Question 1 Detailed Solution

Calculation:
qImage6821ae5650db2c0e03b34c1e
Location of image of A :-

(1/v) - (1/u) = (1/f)

⇒ (1/v) - (1/(v - 30)) = (1/20)

⇒ (1/v) = (1/60)

⇒ v = 60 cm

The magnification m = 2.

Since the size of the object is small with respect to the location, hence

dv = m² du ⇒ dv = 4 × 1 = 4 cm

hi = m h0 ⇒ hi(dy) = 2 × 2 = 4 cm

Therefore, Angle made with principle a×is = -45°

Optics Question 2:

The intensity of transmitted light when a polaroid sheet, placed between two crossed polarization at 22.5° from the polarization axis of one of the polaroids, is (I₀ is the intensity or polarised light after passing through the first polaroid):

  1. \(\frac{I_0}{2}\)
  2. \(\frac{I_0}{4}\)
  3. \(\frac{I_0}{8}\)
  4. \(\frac{I_0}{16}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{I_0}{8}\)

Optics Question 2 Detailed Solution

Calculation:

When light passes through a polaroid at an angle θ from the original polarization direction, the transmitted intensity is given by:

I = I0 × cos²θ

Since there are two polaroids with one in between at 22.5°, the intensity after the first polaroid becomes I0.

After the second polaroid (placed at 22.5° to the first):

I1 = I0 × cos²(22.5°)

After the third polaroid (crossed with the first, i.e., 90° apart, and thus 67.5° from the middle one):

I2 = I1 × cos²(67.5°)

So final intensity = I0 × cos²(22.5°) × cos²(67.5°)

cos(22.5°) ≈ 0.924, cos²(22.5°) ≈ 0.853

cos(67.5°) ≈ 0.383, cos²(67.5°) ≈ 0.146

Final intensity ≈ I0 × 0.853 × 0.146 ≈ I0 × 0.124 ≈ I0 / 8

Correct option: Option 3 (I0 / 8)

Optics Question 3:

In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact.
Then the power of the combination and the total magnification in comparison to the power (p) and magnification (m) for each lens will be, respectively–

  1. 4p and 4m
  2. p⁴ and 4m
  3. 4p and m⁴
  4. p⁴ and m⁴

Answer (Detailed Solution Below)

Option 3 : 4p and m⁴

Optics Question 3 Detailed Solution

Calculation:
For series combination of lens:

peff = p1 + p2 + p3 + p4 = 4p

meff = m1 × m2 × m3 × m4 = m4

Optics Question 4:

What is the angle that is formed between the incident ray and the normal line called?

  1. Angle of deviation
  2. Angle of incidence
  3. Angle of refraction
  4. Angle of reflection

Answer (Detailed Solution Below)

Option 2 : Angle of incidence

Optics Question 4 Detailed Solution

Concept:

When a ray of light strikes a surface, the angle formed between the incoming ray (incident ray) and the normal (an imaginary line perpendicular to the surface) is called the angle of incidence. This is a fundamental concept in the laws of reflection and refraction in optics.

Definition:

Angle of Incidence – The angle between the incident ray and the normal to the surface at the point of incidence.

Optics Question 5:

For Young's double slit experiment, two statements are given below:

Statement I: If screen is moved away from the plane of slits, angular separation of the fringes increases.

Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes increases.

In the light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.

Answer (Detailed Solution Below)

Option 4 : Statement I is false but Statement II is true.

Optics Question 5 Detailed Solution

Concept:

Young's Double Slit Experiment:

  • Young's double-slit experiment demonstrates the interference of light waves from two coherent sources.
  • The angular separation of the fringes is given by \( θ = \frac{λ}{d} \), where \( λ \) is the wavelength of light and \( d \) is the distance between the slits.
  • Angular separation θ : The angle between the central maximum and the first-order maximum.
  • Wavelength λ :
    • Definition: The distance between successive crests of a wave.
    • SI Unit: Meter (m)
    • Dimensional Formula: [L]
    • Formula: \( θ = \frac{λ}{d} \)
  • Distance between slits d:
    • Definition: The distance separating the two slits in the double slit experiment.
    • SI Unit: Meter (m)
    • Dimensional Formula: [L]
  • If the screen is moved away, the linear separation of fringes increases, but the angular separation remains constant since θ" id="MathJax-Element-106-Frame" role="presentation" style="position: relative;" tabindex="0">θ depends only on λ and d.
  • If the wavelength is increased, the angular separation θ increases because θ is directly proportional to λ.

 

Calculation:

Given,

Angular separation \(θ = \frac{λ}{d}\)

For Statement I:

Screen distance changes but \(θ = \frac{λ}{d}\) remains the same.

⇒ Statement I is false.

For Statement II:

If λ increases, θ = λ/d

θ" id="MathJax-Element-107-Frame" role="presentation" style="position: relative;" tabindex="0">θ increases.

⇒ Statement II is true.

∴ Statement I is false but Statement II is true.

Top Optics MCQ Objective Questions

Blue colour of the sky and the reddish appearance of the Sun at the sunrise or sunset is due to

  1. Interference
  2. Reflection
  3. Refraction
  4. Scattering

Answer (Detailed Solution Below)

Option 4 : Scattering

Optics Question 6 Detailed Solution

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The correct answer is Scattering.

Key Points

  • When light passes from one medium to any other medium say air, a glass of water then a part of the light is absorbed by particles of the medium preceded by its subsequent radiation in a particular direction. This phenomenon is termed a scattering of light.
  • Rayleigh's law of scattering:-
    • According to Rayleigh's law of scattering, the intensity of light of wavelength λ present in the scattered light is inversely proportional to the fourth power of λ, provided the size of the scattering particles is much smaller than λ. Mathematically,
    • \(I \propto \frac{1}{\lambda^4 }\)
  • Thus the scattered intensity is maximum for a shorter wavelength.

Additional Information

  • The molecules of air and other fine particles in the atmosphere have a size smaller than the wavelength of visible light.
  • These are more effective in scattering light of shorter wavelengths at the blue end than the light of longer wavelengths at the red end.
  • The red light has a wavelength of about 1.8 times greater than blue light.
  • Thus, when sunlight passes through the atmosphere, the fine particles in the air scatter the blue colour (shorter wavelengths) more strongly than red
  • The scattered blue light enters our eyes
  • The red colour of the sun at sunrise is due to the scattering of light

Sky-is-made-up-of-air-and-light 

The power of lens of focal length 1 cm is

  1. 1 D
  2. 10 D
  3. 100 D
  4. 0.01 D

Answer (Detailed Solution Below)

Option 3 : 100 D

Optics Question 7 Detailed Solution

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CONCEPT:

  • Power of the Lens: The lens power is the inverse of the focal length in meters.
    • The physical unit for lens power is 1/meter which is called dioptre (D).

Power (P) = 1/f 

Where f is the focal length of the lens

CALCULATION:

Given that:

Focal length of the lens (f) = 1 cm = 1/100 = 0.01 m

So the power of the lens, \(p\; = \;\frac{1}{{f}}\; = \;\frac{1}{0.01}\; = \;100\;D\)

Find the magnification of the convex lens, if an object is placed 30 cm from a convex lens which has a focal length of 15 cm.

  1. +1
  2. +0.5
  3. –0.5
  4. -1

Answer (Detailed Solution Below)

Option 4 : -1

Optics Question 8 Detailed Solution

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CONCEPT:

  • Lens formula: The expression which shows the relation between object distance (u), image distance (v), and focal length (f) is called lens formula.

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

Linear magnification (m):

  • It is defined as the ratio of the height of the image (hi) to the height of the object (ho).

i.e. \(m = \frac{{{h_i}}}{{{h_o}}}\)

  • The ratio of image distance to the object distance is called linear magnification.

i.e. \(m = \frac{{image\;distance\;\left( v \right)}}{{object\;distance\;\left( u \right)}} = \frac{v}{u}\)

CALCULATION:

Given: u = - 30 cm and f = 15 cm

Lens formula

\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)

\(\therefore \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{{15}} - \frac{1}{{30}} = \frac{{2 - 1}}{{30}} = \frac{1}{{30}}\;cm\)

v = 30 cm

Linear magnification (m)

\(m = \frac{v}{u}\)

\(\Rightarrow m = \frac{{30}}{{ - 30}} = - 1\)

What is the nature of the image formed due to concave mirror when object is placed beyond centre of curvature ?

  1. Real and inverted
  2. Virtual and erect
  3. Real and erect
  4. Virtual and inverted

Answer (Detailed Solution Below)

Option 1 : Real and inverted

Optics Question 9 Detailed Solution

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The correct answer is Real and inverted.

  • The image formed by the concave mirror when the object is placed beyond the center of curvature is Real and Inverted.

Key Points

  • Concave Mirror:
    • A spherical mirror, whose reflecting surface is curved inwards, that is, faces towards the center of the sphere, is called a concave mirror.
    • Concave mirrors are commonly used in  Torches, Search-lights, Vehicle headlights, Shaving mirrors, Microscopes, and Telescopes.
    • Image formation by a concave mirror for different positions of the object:
Position of the object Position of the image Size of the image Nature of the image
At infinity At the focus F Highly diminished  Real and inverted
Beyond C Between F and C Diminished  Real and inverted
At C At C Same size  Real and inverted
Between C and F Beyond C Enlarged  Real and inverted
At F At infinity Highly enlarged  Real and inverted
Between P and F Behind the mirror Enlarged Virtual and erect

concave

_________ are used by dentists to see an enlarged image of the teeth.

  1. Concave mirrors
  2. Convex mirrors
  3. Bifocal mirrors
  4. Plain mirrors

Answer (Detailed Solution Below)

Option 1 : Concave mirrors

Optics Question 10 Detailed Solution

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The correct answer is Concave mirrors.

Key Points 

  • Concave mirrors are spherical mirror in which the reflecting surface curved inwards.
  • A concave mirror forms a virtual or real image depending on the position of the object.
  • Concave mirrors are used by dentists to see an enlarged image of the teeth.
  • Other important uses of Concave mirrors are:
    • ​Used in torches.
    • Used in solar cooker.
    • Used in search-lights and vehicles headlights to get powerful parallel beams of light.
    • Used as shaving mirrors to see a larger image of the face.
  • Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Additional Information 

  • Convex mirrors are the spherical mirrors in which the reflecting surface curved outwards.
  • The important uses of a convex mirror are:
    • Used as rear-view mirrors in vehicles.
    • Used for Security purposes.
  • Plane mirrors are used in Periscopes and kaleidoscopes.

Image formed by Plane Mirror is always ________.

  1. Virtual and Erect
  2. Real and Inverted
  3. Real and Erect
  4. Virtual and Inverted

Answer (Detailed Solution Below)

Option 1 : Virtual and Erect

Optics Question 11 Detailed Solution

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CONCEPT:

  • Plane Mirror: A plane mirror is a mirror with a flat (planar) reflective surface.

The characteristics of an image formed in a plane mirror:

F1 J.K 2.6.20 Pallavi D8

  • The image formed by the plane mirror is virtual and erect i.e. image cannot be projected or focused on a screen.
  • The distance of the image ‘behind’ the mirror is the same as the distance of the object in front of the mirror.
  • The size of the image formed is the same as the size of the object.
  • The image is laterally inverted, i.e. left hand appears to be right hand when seen from the plane mirror.

F1 J.K 2.6.20 Pallavi D9

  • If the object moves towards (or away from) the mirror at a certain rate, the image also moves towards (or away from) the mirror at the same rate.

EXPLANATION:

From the above discussion, we can say that,

  • The image formed by Plane Mirror is always Virtual and Erect. So option 1 is correct.
  • A convex lens and a concave mirror forms both real and virtual images.
  • A concave lens and a convex mirror can form only virtual images.

The image formed by concave mirror is real, inverted and of the same size as that of the object. The position of the object should be

  1. at the focus
  2.  at the centre of curvature
  3. between the focus and centre of curvature
  4. beyond the centre of curvature

Answer (Detailed Solution Below)

Option 2 :  at the centre of curvature

Optics Question 12 Detailed Solution

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The correct answer is at the centre of the curvature.

Key Points

Position of the Object at the Centre of Curvature: 
When an object is placed at the center of curvature (C) of a concave mirror, the object is essentially at a distance equal to the radius of curvature from the mirror's pole.
Formation of Image:

  • In this scenario, the reflected rays from the object converge and meet at the center of curvature.
  • The image formed is real because the reflected rays actually converge at a point.
  • The image is inverted because the reflected rays cross over at the center of curvature, resulting in an inverted orientation.
  • The image is of the same size as the object because the distance from the object to the mirror is the same as the distance from the image to the mirror.
  • If an object is placed at the centre of curvature of a concave mirror, then the image formed is real, inverted, and of the same size as that of the object.

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An object is placed at 20 cm in front of a concave mirror produces three time magnified real image. What is focal length of the concave mirror?

  1. 15 cm
  2. 6.6 cm
  3. 10 cm
  4. 7.5 cm

Answer (Detailed Solution Below)

Option 1 : 15 cm

Optics Question 13 Detailed Solution

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CONCEPT:

Concave mirror: The mirror in which the rays converge after falling on it is known as the concave mirror.

  • Concave mirrors are also known as a converging mirror.
  • The focal length of a concave mirror is negative according to the sign convention.

Magnification: In a concave mirror, the magnification is the ratio of the height of the image to the height of the object.

When the image is real, the magnification will be negative because the real image is inverted.

  • When the image is virtual, the magnification will be positive because the virtual image is erect.

 \(m = \frac{-v}{u}\)

where m is magnificent, v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

  • Mirror Formula: The following formula is known as the mirror formula:

\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

where f is focal length v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

CALCULATION:

Given that u = -20 cm and m = -3.

\(m = \frac{-v}{u} \)

\(-3=\frac{-v}{-20}\)

v = -60 cm

Mirror formula \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

\(\frac{1}{f}=\frac{1}{-20}+\frac{1}{-60}\)

f = -60 / 4= -15 cm

So the correct answer is option 1.

The magnification of a compound microscope is 20. The focal length of the eyepiece is 5 cm and the image is formed at near point (25 cm). The magnification of the objective lens is 

  1. 5.8
  2. 6.78
  3. 3.33
  4. 4.1

Answer (Detailed Solution Below)

Option 3 : 3.33

Optics Question 14 Detailed Solution

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CONCEPT:

Compound microscope: 

  • It is used for much larger magnifications.
  • Two lenses are used with one lens compounding the effect of the other.

F1 Prabhu 17.6.21 Pallavi D3

where fo is the focal length of the objective lens,  f is the focal length of the eyepiece, h is the object height, h' is the size of the first image

  • A compound microscope uses two lenses i.e., objective lens and eyepiece
    • Objective lens: lens nearest to the object forming a real, inverted, and magnified image of the object.
    • Eyepiece: produces a final image that is enlarged and virtual.

CALCULATION:

Given: magnification of compound microscope(m) = 20, focal length of eye piece (fe) = 5 cm and image is formed at near point (v= D) = 25 cm.

  • Linear magnification due to eyepiece:

\(⇒ m_{e} = 1+\frac{D}{f} = 1+ (\frac {25}{5} )= 6\)

  • Total magnification is given by:

⇒ m = m× me,

where mo is the magnification due to the objective lens.

  • Magnification due to the objective lens is given by:

\(⇒ m_{o} = \frac {m}{m_{e}} = \frac {20}{6} = 3.33\)

  • Hence option 3) is correct.

The refractive index for a glass is 1.62. The sine of the critical angle for total internal refection at a glass-air interface is: 

  1. 0.40
  2. 0.62
  3. 0.74
  4. 1.00

Answer (Detailed Solution Below)

Option 2 : 0.62

Optics Question 15 Detailed Solution

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Concept:

Refraction of Light: When a ray of light is traveling from one transparent medium to another, it bends its path. This phenomenon is called refraction. 

When the light changes its medium, its speed, and wavelength changes.

F1 J.K 8.5.20 Pallavi D3

Where i is the angle of incidence, r is the angle of refraction.

Refractive Index: 

The ratio of the speed of light in the vacuum to the speed of light in a given transparent medium is called the refractive index of the medium.

  • When light travels from a medium with a higher refractive index to a lower one, then it bends away from normal. The angle of refraction is more than the angle of incidence.
  • When light travels from a medium with a lower refractive index to a higher one, then it bends toward normal. The angle of refraction is less than the angle of incidence.

 

Snells Law of Refraction: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for two transparent medium.

\(\frac{sin\; i}{sin\; r} = \;\frac{n_2}{n_1}\)

n2 is the medium in which light is entering, n1 is the initial medium of light

When the ray of light is traveling from the medium with a higher refractive index to a lower one, for example, water to air, and the angle of refraction formed is 90 ° then the angle of incidence at that point is critical angle.

F1 J.K 8.5.20 Pallavi D4

When medium 2 is air having refractive index n2 = 1, and medium 1 have refractive index n1 =  μ, then critical angle θc can be represented by 

\(\frac{sin\; θ_c}{sin \;90 °} = \frac{1}{μ}\)

\(\implies sin\; θ_c= \frac{1}{μ}\)

  • The sine of critical angle is reciprocal of refractive index.
  • When the incidence angle is more than the critical angle, then the ray of light reflects back in the same medium. This phenomenon is called Total Internal Reflection.

 

Calculation:

Given refractive index μ = 1. 62

\(sin\; θ_c= \frac{1}{μ}\)

\(sin\; θ _c= \frac{1}{1.62}\)

sin θc = 0. 62

So, 0.62 is the correct option.

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