Refractive index MCQ Quiz - Objective Question with Answer for Refractive index - Download Free PDF

Last updated on May 30, 2025

Latest Refractive index MCQ Objective Questions

Refractive index Question 1:

The minimum deviation produced by a hollow prism filled with a certain liquid is found to be 30°. The light ray is also found to be refracted at an angle of 30°. Then the refractive index of the liquid is

  1. √2
  2. √3
  3. \(\sqrt\frac{3}{2}\)
  4. \(\frac{3}{2}\)

Answer (Detailed Solution Below)

Option 1 : √2

Refractive index Question 1 Detailed Solution

Concept:

In the case of a hollow prism filled with a liquid, the refractive index (n) of the liquid can be determined using the relationship between the angle of minimum deviation (D) and the refractive index of the material filling the prism. The formula for the refractive index of the liquid is:

n = sin((D + A) / 2) / sin(A / 2),

where:

  • D is the minimum deviation angle,
  • A is the prism angle,
  • n is the refractive index of the liquid.

Additionally, the refractive angle inside the liquid can also help determine the refractive index using the Snell's law:

n = sin(i) / sin(r),

where i is the angle of incidence and r is the angle of refraction inside the liquid.

Calculation:

Given:

  • Minimum deviation D = 30°
  • Refracted angle inside the liquid r = 30°

The angle of refraction inside the liquid is equal to the angle of incidence (since the light ray is refracted symmetrically inside the liquid). Thus, we can directly use the relationship from Snell's law for a refracted angle of 30°.

Applying the formula:

n = sin(30°) / sin(30°) = 1 / √2

∴ The refractive index of the liquid is n = √2. Option 1) is correct.

Refractive index Question 2:

When a ray of light enters from medium A to medium B, its speed decreases and on further entering into medium C, its speed increases.
Which of the following conclusions can be drawn from this ?

  1. Refractive index of A is more than B and C
  2. Refractive index of A is less than B but equal to C
  3. Refractive index of A and C are less than B
  4. Refractive index of A, B, C cannot be compared from above information

Answer (Detailed Solution Below)

Option 3 : Refractive index of A and C are less than B

Refractive index Question 2 Detailed Solution

The correct answer is: Option 3) Refractive index of A and C are less than B

Explanation:

When a ray of light passes from one medium to another, its speed changes according to the refractive index of the media. The refractive index (n) is given by:

n = c / v, where:

c is the speed of light in a vacuum

v is the speed of light in the medium

From the question, we know that the speed of light decreases when it enters medium B from medium A, and it increases when entering medium C from medium B.

If the speed decreases when entering medium B, it means the refractive index of B is greater than that of A because a higher refractive index corresponds to a lower speed of light in the medium.

If the speed increases when entering medium C, it means the refractive index of C is less than that of B, because a lower refractive index corresponds to a higher speed of light in the medium.

Refractive index Question 3:

A transparent glass cube of length 24 cm has a small air bubble trapped inside. When seen normally through one surface from air outside, its apparent distance is 10 cm from the surface. When seen normally from opposite surface, its apparent distance is 6 cm. The distance of the air bubble from first surface is

  1. 15 cm
  2. 14 cm
  3. 12 cm
  4. 8 cm

Answer (Detailed Solution Below)

Option 1 : 15 cm

Refractive index Question 3 Detailed Solution

Answer : 1

Solution :

Given : Length of cube = 12 cm

qImage670a207bfa1f5634e63475e5

\(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{l_1}{\mathrm{~h}_1}=\frac{24-l_1}{\mathrm{~h}_2}\)

putting h1 = 10 cm and h2 = 6 cm into (i), we get \(\frac{l_1}{10}=\frac{24-l_1}{6}\)

6l1 = 240 - 10l1

16l1 = 240

∴ l1 = 15 cm

Refractive index Question 4:

In finding out the refractive index of the glass slab the following observations were made through traveling microscope 50 vernier scale division = 49 MSD; 20 divisions on the main scale in each cm For mark on paper 

MSR = 8.45 cm, VC = 26

For mark on paper seen through slab

MSR = 7.12 cm, VC = 41

For powder particle on the top surface of the glass slab 

MSR = 4.05 cm, VC = 1

(MSR = Main Scale Reading, VC = Vernier Coincidence)

1.42Refractive index of the glass slab is:

  1. 1.42
  2. 1.52
  3. 1.24
  4. 1.35

Answer (Detailed Solution Below)

Option 1 : 1.42

Refractive index Question 4 Detailed Solution

Calculation: 

1 MSD = \(\frac{1 \mathrm{~cm}}{20}\)  0.05 cm

1 VSD = \(\frac{49}{50}\) MSD = \(\frac{49}{50}\) × 0.05 cm = 0.049 cm 

LC = 1MSD – 1VSD = 0.001 cm

For mark on paper, L1 = 8.45 cm + 26 × 0.001 cm = 84.76 mm

For mark on paper through slab, L2 = 7.12 cm + 41× 0.001 cm = 71.61 mm 

For powder particle on top surface, ZE = 4.05 cm + 1 × 0.001 cm = 40.51 mm

⇒ actual L1 = 84.76 – 40.51 = 44.25 mm

actual L2 = 71.61 – 40.51 = 31.10 mm 

L\(\frac{\mathrm{L}_1}{μ}\)

 μ = \(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\) = \(\frac{44.25}{31.10}\) = 1.42

∴ The Correct answer is Option (1): 1.42 

Refractive index Question 5:

A ray of light propagating through medium A enters another medium B such that the angle of incidence and angle of refraction are α and β. Given that sin α = \(({1\over x})\) and sin β = \(({1\over y})\), the refractive index of medium A with respect to medium B is

  1. \(x\over y\)
  2. \(y\over x\)
  3. xy
  4. \(1\over xy\)
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : \(x\over y\)

Refractive index Question 5 Detailed Solution

Concept:

Refractive index:

  • It is the ratio between the speed of light in air to the speed in a medium.
  • Formula, refractive index, \(μ =\frac{c}{v}\)
  • In terms of the wavelength, it is equal to the ratio of the wavelength of light in air to the wavelength of light in the medium.
  • Formula, \(μ =\frac{λ_0}{λ}\) where, λ0 = wavelength in the air, λ = wavelength in the medium.

Snell's law:

  • Snell’s law gives the degree of refraction and the relation between the angle of incidence, the angle of refraction, and the refractive indices of a given pair of media. 
  • Snell’s law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for the given pair of media”. Snell’s law formula is expressed as:

  • Formula, n1sinθ1 = n2 sinθ2  

Calculation:

Given,

sin α = \(({1\over x})\) and sin β = \(({1\over y})\), 

α = angle of incidence, β = angle of refraction

Snell's law, n1sinθ1 = n2 sinθ2  

The refractive index of medium A with respect to medium B is,

\(n=\frac{n_1}{n_2}=\frac{sin β }{sin\alpha}\)

\(n=\frac{1/y }{1/x}=\frac{x}{y}\)

Top Refractive index MCQ Objective Questions

The refractive index of glass is 1.5 for light whose wavelength in a vacuum is 6000 Å. The wavelength of this light when it passes through glass is

  1. 4000 Å
  2. 6000 Å
  3. 9000 Å
  4. 15000 Å

Answer (Detailed Solution Below)

Option 1 : 4000 Å

Refractive index Question 6 Detailed Solution

Download Solution PDF

Concept:

Refractive index:

  • It is the ratio between the speed of light in air to the speed in a medium.
  • Formula, refractive index, \(μ =\frac{c}{v}\)
  • In terms of the wavelength, it is equal to the ratio of the wavelength of light in air to the wavelength of light in the medium.
  • Formula, \(μ =\frac{λ_0}{λ}\) where, λ0 = wavelength in the air, λ = wavelength in the medium.

Calculation:

Given, 

The refractive index, μ = 1.5

The wavelength in the vacuum, λ0 = 6000 Å

We  know that, \(μ =\frac{λ_0}{λ}\)

\(\lambda =\frac{λ_0}{\mu}\)

 \(\lambda =\frac{6000}{1.5}\)

λ = 4000 Å

A ray of light propagating through medium A enters another medium B such that the angle of incidence and angle of refraction are α and β. Given that sin α = \(({1\over x})\) and sin β = \(({1\over y})\), the refractive index of medium A with respect to medium B is

  1. \(x\over y\)
  2. \(y\over x\)
  3. xy
  4. \(1\over xy\)

Answer (Detailed Solution Below)

Option 1 : \(x\over y\)

Refractive index Question 7 Detailed Solution

Download Solution PDF

Concept:

Refractive index:

  • It is the ratio between the speed of light in air to the speed in a medium.
  • Formula, refractive index, \(μ =\frac{c}{v}\)
  • In terms of the wavelength, it is equal to the ratio of the wavelength of light in air to the wavelength of light in the medium.
  • Formula, \(μ =\frac{λ_0}{λ}\) where, λ0 = wavelength in the air, λ = wavelength in the medium.

Snell's law:

  • Snell’s law gives the degree of refraction and the relation between the angle of incidence, the angle of refraction, and the refractive indices of a given pair of media. 
  • Snell’s law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for the given pair of media”. Snell’s law formula is expressed as:

  • Formula, n1sinθ1 = n2 sinθ2  

Calculation:

Given,

sin α = \(({1\over x})\) and sin β = \(({1\over y})\), 

α = angle of incidence, β = angle of refraction

Snell's law, n1sinθ1 = n2 sinθ2  

The refractive index of medium A with respect to medium B is,

\(n=\frac{n_1}{n_2}=\frac{sin β }{sin\alpha}\)

\(n=\frac{1/y }{1/x}=\frac{x}{y}\)

The velocity of light in water of refractive index \(\frac{4}{3}\) in ms−1 is

  1. 1.33 × 108
  2. 2.25 × 108
  3. 3 × 108
  4. 4 × 108

Answer (Detailed Solution Below)

Option 2 : 2.25 × 108

Refractive index Question 8 Detailed Solution

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Concept:

Refractive Index (ηm):

The refractive index of a transparent material is defined as the ratio of the speed of light in the vacuum to that of a medium.

\(η_m= \frac{c}{v}\)

where, c = speed of light in vacuum, v = speed of light in the medium.

Calculation:

Given:

ηm = 4/3, c = 3 × 108 m/s.

\(η_m= \frac{c}{v}\)

\(v= \frac{3× 10^8× 3}{4}\)

v = 2.25 × 108 m/s.

Refractive index Question 9:

The refractive index of glass is 1.5 for light whose wavelength in a vacuum is 6000 Å. The wavelength of this light when it passes through glass is

  1. 4000 Å
  2. 6000 Å
  3. 9000 Å
  4. 15000 Å

Answer (Detailed Solution Below)

Option 1 : 4000 Å

Refractive index Question 9 Detailed Solution

Concept:

Refractive index:

  • It is the ratio between the speed of light in air to the speed in a medium.
  • Formula, refractive index, \(μ =\frac{c}{v}\)
  • In terms of the wavelength, it is equal to the ratio of the wavelength of light in air to the wavelength of light in the medium.
  • Formula, \(μ =\frac{λ_0}{λ}\) where, λ0 = wavelength in the air, λ = wavelength in the medium.

Calculation:

Given, 

The refractive index, μ = 1.5

The wavelength in the vacuum, λ0 = 6000 Å

We  know that, \(μ =\frac{λ_0}{λ}\)

\(\lambda =\frac{λ_0}{\mu}\)

 \(\lambda =\frac{6000}{1.5}\)

λ = 4000 Å

Refractive index Question 10:

The value of absolute refractive index of a medium is always:

  1. less than 1
  2. equal to 1
  3. equal to 0
  4. more than 1

Answer (Detailed Solution Below)

Option 4 : more than 1

Refractive index Question 10 Detailed Solution

The correct answer is more than 1.Key Points

  • Refractive index is a measure of how much a ray of light bends when it passes through a medium.
  • Absolute refractive index is the ratio of the speed of light in a vacuum to the speed of light in the medium.
  • The value of absolute refractive index of a medium is always more than 1.

Additional Information

  • The characteristics of a medium affect the speed of light within it.
  • The optical density of the medium affects the speed of electromagnetic waves.
  • The atoms' propensity to replenish the electromagnetic energy they have received is known as optical density.
  • The speed of light decreases with increasing optical density.
  • The refractive index is one such measure of a medium's optical density.
  • The refractive index of a medium depends on the density and composition of the medium.
  • When the refractive index of a medium is more than 1, it means that the speed of light is slower in that medium than in a vacuum.

Refractive index Question 11:

A ray of light propagating through medium A enters another medium B such that the angle of incidence and angle of refraction are α and β. Given that sin α = \(({1\over x})\) and sin β = \(({1\over y})\), the refractive index of medium A with respect to medium B is

  1. \(x\over y\)
  2. \(y\over x\)
  3. xy
  4. \(1\over xy\)

Answer (Detailed Solution Below)

Option 1 : \(x\over y\)

Refractive index Question 11 Detailed Solution

Concept:

Refractive index:

  • It is the ratio between the speed of light in air to the speed in a medium.
  • Formula, refractive index, \(μ =\frac{c}{v}\)
  • In terms of the wavelength, it is equal to the ratio of the wavelength of light in air to the wavelength of light in the medium.
  • Formula, \(μ =\frac{λ_0}{λ}\) where, λ0 = wavelength in the air, λ = wavelength in the medium.

Snell's law:

  • Snell’s law gives the degree of refraction and the relation between the angle of incidence, the angle of refraction, and the refractive indices of a given pair of media. 
  • Snell’s law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for the given pair of media”. Snell’s law formula is expressed as:

  • Formula, n1sinθ1 = n2 sinθ2  

Calculation:

Given,

sin α = \(({1\over x})\) and sin β = \(({1\over y})\), 

α = angle of incidence, β = angle of refraction

Snell's law, n1sinθ1 = n2 sinθ2  

The refractive index of medium A with respect to medium B is,

\(n=\frac{n_1}{n_2}=\frac{sin β }{sin\alpha}\)

\(n=\frac{1/y }{1/x}=\frac{x}{y}\)

Refractive index Question 12:

The velocity of light in water of refractive index \(\frac{4}{3}\) in ms−1 is

  1. 1.33 × 108
  2. 2.25 × 108
  3. 3 × 108
  4. 4 × 108

Answer (Detailed Solution Below)

Option 2 : 2.25 × 108

Refractive index Question 12 Detailed Solution

Concept:

Refractive Index (ηm):

The refractive index of a transparent material is defined as the ratio of the speed of light in the vacuum to that of a medium.

\(η_m= \frac{c}{v}\)

where, c = speed of light in vacuum, v = speed of light in the medium.

Calculation:

Given:

ηm = 4/3, c = 3 × 108 m/s.

\(η_m= \frac{c}{v}\)

\(v= \frac{3× 10^8× 3}{4}\)

v = 2.25 × 108 m/s.

Refractive index Question 13:

When a ray of light enters from medium A to medium B, its speed decreases and on further entering into medium C, its speed increases.
Which of the following conclusions can be drawn from this ?

  1. Refractive index of A is more than B and C
  2. Refractive index of A is less than B but equal to C
  3. Refractive index of A and C are less than B
  4. Refractive index of A, B, C cannot be compared from above information

Answer (Detailed Solution Below)

Option 3 : Refractive index of A and C are less than B

Refractive index Question 13 Detailed Solution

The correct answer is: Option 3) Refractive index of A and C are less than B

Explanation:

When a ray of light passes from one medium to another, its speed changes according to the refractive index of the media. The refractive index (n) is given by:

n = c / v, where:

c is the speed of light in a vacuum

v is the speed of light in the medium

From the question, we know that the speed of light decreases when it enters medium B from medium A, and it increases when entering medium C from medium B.

If the speed decreases when entering medium B, it means the refractive index of B is greater than that of A because a higher refractive index corresponds to a lower speed of light in the medium.

If the speed increases when entering medium C, it means the refractive index of C is less than that of B, because a lower refractive index corresponds to a higher speed of light in the medium.

Refractive index Question 14:

A ray of light propagating through medium A enters another medium B such that the angle of incidence and angle of refraction are α and β. Given that sin α = \(({1\over x})\) and sin β = \(({1\over y})\), the refractive index of medium A with respect to medium B is

  1. \(x\over y\)
  2. \(y\over x\)
  3. xy
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : \(x\over y\)

Refractive index Question 14 Detailed Solution

Concept:

Refractive index:

  • It is the ratio between the speed of light in air to the speed in a medium.
  • Formula, refractive index, \(μ =\frac{c}{v}\)
  • In terms of the wavelength, it is equal to the ratio of the wavelength of light in air to the wavelength of light in the medium.
  • Formula, \(μ =\frac{λ_0}{λ}\) where, λ0 = wavelength in the air, λ = wavelength in the medium.

Snell's law:

  • Snell’s law gives the degree of refraction and the relation between the angle of incidence, the angle of refraction, and the refractive indices of a given pair of media. 
  • Snell’s law is defined as “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for the given pair of media”. Snell’s law formula is expressed as:

  • Formula, n1sinθ1 = n2 sinθ2  

Calculation:

Given,

sin α = \(({1\over x})\) and sin β = \(({1\over y})\), 

α = angle of incidence, β = angle of refraction

Snell's law, n1sinθ1 = n2 sinθ2  

The refractive index of medium A with respect to medium B is,

\(n=\frac{n_1}{n_2}=\frac{sin β }{sin\alpha}\)

\(n=\frac{1/y }{1/x}=\frac{x}{y}\)

Refractive index Question 15:

Light, from a 520 nm monochromatic source in air, is incident on a rectangular slab of quartz (refractive index μ = 1.56) at an angle of 30°. The wavelength of light, refracted within the quartz, is close to

  1. 173 nm
  2. 333 nm
  3. 520 nm
  4. 811 nm

Answer (Detailed Solution Below)

Option 2 : 333 nm

Refractive index Question 15 Detailed Solution

Concept:

The frequency of light does not depend on the property of the medium in which it is traveling. Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in the air.

Calculation:'

Light from a 520 nm monochromatic source in air is incident on a rectangular slab of quartz with a refractive index (μ) of 1.56 at an angle of 30°.

The wavelength of light refracted within the quartz is calculated as follows:

λmedium = λair / μ

Where,

  • λair = 520 nm
  • μ = 1.56

λquartz = 520 nm / 1.56 ≈ 333.33 nm

Thus, the wavelength of light within the quartz is approximately 333 nm.

The correct answer is option (1).

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