Polarization by Reflection MCQ Quiz - Objective Question with Answer for Polarization by Reflection - Download Free PDF

Calculation:

θ_B = tan^-1(n) = tan^-1(√3) = 60^°

r = ∠BCD = 60^°, i = ∠BCE = 30^°

AE = 3R / 4, AD = R ⇒ sin(θ) = AE / AD = 3 / 4 = 0.75

For light to be fully polarized at point O, it must be incident at Brewster’s angle:

For refraction at C (glass to air interface):

Since CE is the angle bisector of ∠BCE, it is perpendicular to BD.

In triangle geometry:

Final Answer: sin(θ) = 0.75

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.

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.

Answer : 1

Solution :

Given : Length of cube = 12 cm

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

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

Concept: 

To solve this problem, we can use Malus's Law, which states that the intensity of light passing through a polarizer-analyzer set is given by:

$I=I_{0}Co{s}^2\phi$

Calculation:

So, 0.1Io = Iocos2φ

⇒ cosφ = √0.1

⇒ cosφ = 0.316

Since, cos φ < cos 45o therefore, φ > 45° If the light is passing at 90° from the plane of polaroid, than its intensity will be zero.

Then, θ = 90° − φ therefore, θ will be less than 45°. So, the only option matching is option d which is 18.4°

∴ The Correct answer is Option (4): 18.4°

Concept:

Refractive index:

• It is the ratio between the speed of light in air to the speed in a medium.
• Formula, refractive index, $\mu =\frac{c}{v}$
• 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, $\mu =\frac{{\lambda }_0}{\lambda }$ 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, $\mu =\frac{{\lambda }_0}{\lambda }$

$\lambda =\frac{{\lambda }_0}{\mu }$

 $\lambda =\frac{6000}{1.5}$

λ = 4000 Å

Concept:

Refractive index:

• It is the ratio between the speed of light in air to the speed in a medium.
• Formula, refractive index, $\mu =\frac{c}{v}$
• 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, $\mu =\frac{{\lambda }_0}{\lambda }$ 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 = $(\frac{1}{x})$ and sin β = $(\frac{1}{y})$, 

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,

$n=\frac{n_1}{n_2}=\frac{sin\beta }{sin\alpha }$

$n=\frac{1/y}{1/x}=\frac{x}{y}$

Concept:

• Polarized waves are light waves in which the vibrations occur in a single plane.
•  Polarization of light by reflection,
  • If a beam of light strikes an interface so that there is a 90° angle between the reflected and refracted beams, the reflected beam will be linearly polarized.
• The Brewster angle, the angle of incidence required to produce a linearly polarized reflected beam, is given by,
• tan θ_B = $\frac{{\mu }_2}{{\mu }_1}$ --- (1)

• Speed of light $\frac{v_1}{v_2}=\frac{{\mu }_2}{{\mu }_1}$ --- (2)
• Where, v₁ = speed of light in medium 1, v₂ = speed of light in medium 2,
• Speed of light = 3 × 10⁸ m/s

Calculation:

Given angle of incidence, θ_B = 60, 

From equation 1,

$\frac{{\mu }_2}{{\mu }_1}$ = tan θ_B = tan 60° = √3 

Now from equation 2, and also v₁ = 3 × 10⁸ m/s

$\frac{v_1}{v_2}=\frac{{\mu }_2}{{\mu }_1}$

$\Rightarrow \frac{3\times {10}^8}{v_2}=\surd 3$

⇒ v₂ = √3 × 10⁸ m/s 

The velocity of refracted ray inside the material in m/s is √3 × 10⁸ m/s 

Concept:

Refractive Index (η_m):

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.

${\eta }_m=\frac{c}{v}$

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

Calculation:

Given:

η_m = 4/3, c = 3 × 10⁸ m/s.

${\eta }_m=\frac{c}{v}$

$v=\frac{3\times {10}^8\times 3}{4}$

v = 2.25 × 108 m/s.

Concept: 

To solve this problem, we can use Malus's Law, which states that the intensity of light passing through a polarizer-analyzer set is given by:

$I=I_{0}Co{s}^2\phi$

Calculation:

So, 0.1Io = Iocos2φ

⇒ cosφ = √0.1

⇒ cosφ = 0.316

Since, cos φ < cos 45o therefore, φ > 45° If the light is passing at 90° from the plane of polaroid, than its intensity will be zero.

Then, θ = 90° − φ therefore, θ will be less than 45°. So, the only option matching is option d which is 18.4°

∴ The Correct answer is Option (4): 18.4°

Concept:

Refractive index:

• It is the ratio between the speed of light in air to the speed in a medium.
• Formula, refractive index, $\mu =\frac{c}{v}$
• 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, $\mu =\frac{{\lambda }_0}{\lambda }$ 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, $\mu =\frac{{\lambda }_0}{\lambda }$

$\lambda =\frac{{\lambda }_0}{\mu }$

 $\lambda =\frac{6000}{1.5}$

λ = 4000 Å

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.

Concept:

Refractive index:

• It is the ratio between the speed of light in air to the speed in a medium.
• Formula, refractive index, $\mu =\frac{c}{v}$
• 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, $\mu =\frac{{\lambda }_0}{\lambda }$ 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 = $(\frac{1}{x})$ and sin β = $(\frac{1}{y})$, 

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,

$n=\frac{n_1}{n_2}=\frac{sin\beta }{sin\alpha }$

$n=\frac{1/y}{1/x}=\frac{x}{y}$

Concept:

• Polarized waves are light waves in which the vibrations occur in a single plane.
•  Polarization of light by reflection,
  • If a beam of light strikes an interface so that there is a 90° angle between the reflected and refracted beams, the reflected beam will be linearly polarized.
• The Brewster angle, the angle of incidence required to produce a linearly polarized reflected beam, is given by,
• tan θ_B = $\frac{{\mu }_2}{{\mu }_1}$ --- (1)

• Speed of light $\frac{v_1}{v_2}=\frac{{\mu }_2}{{\mu }_1}$ --- (2)
• Where, v₁ = speed of light in medium 1, v₂ = speed of light in medium 2,
• Speed of light = 3 × 10⁸ m/s

Calculation:

Given angle of incidence, θ_B = 60, 

From equation 1,

$\frac{{\mu }_2}{{\mu }_1}$ = tan θ_B = tan 60° = √3 

Now from equation 2, and also v₁ = 3 × 10⁸ m/s

$\frac{v_1}{v_2}=\frac{{\mu }_2}{{\mu }_1}$

$\Rightarrow \frac{3\times {10}^8}{v_2}=\surd 3$

⇒ v₂ = √3 × 10⁸ m/s 

The velocity of refracted ray inside the material in m/s is √3 × 10⁸ m/s 

Concept:

Refractive Index (η_m):

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.

${\eta }_m=\frac{c}{v}$

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

Calculation:

Given:

η_m = 4/3, c = 3 × 10⁸ m/s.

${\eta }_m=\frac{c}{v}$

$v=\frac{3\times {10}^8\times 3}{4}$

v = 2.25 × 108 m/s.