Question

Time taken by light to travel in two different materials A and B of refractive indices μ_A and μ_B of same thickness is t₁ and t₂ respectively. If t₂ – t₁ = 5 × 10^–10 s and the ratio of μA to μB is 1 : 2. Then, the thickness of material, in meter is: (Given v_A and v_B are velocities of light in A and B materials, respectively.)

• A. 5 × 10^-10 v_A m
• B. 5 × 10^-10 m
• C. 1.5 × 10^-10 m
• D. 5 × 10^-10 v_B m

Answer 1

The Correct Option is A

To find the thickness of the material, we first need to understand the relationship between refractive index, velocity of light in the material, and the time it takes for light to travel through the material.

Step 1: Understanding the Refractive Index

The refractive index \( \mu \) of a material is defined as the ratio of the speed of light in vacuum \( c \) to the speed of light in the material \( v \). Therefore,

\mu = \frac{c}{v}

Step 2: Understand the Relationship Between Time, Distance, and Velocity

The thickness of the material is the same for both materials A and B. Hence, if \( d \) is the thickness, the time taken for light to travel through the materials can be given by:

• For material A: t_1 = \frac{d}{v_A}
• For material B: t_2 = \frac{d}{v_B}

Step 3: Relate the Time Difference to Refractive Index and Velocity

We know from the problem statement that:

t_2 - t_1 = 5 \times 10^{-10} \, \text{s}

Substituting for \( t_1 \) and \( t_2 \) from above, we get:

\frac{d}{v_B} - \frac{d}{v_A} = 5 \times 10^{-10} \, \text{s}

Factoring out \( d \), we have:

d \left( \frac{1}{v_B} - \frac{1}{v_A} \right) = 5 \times 10^{-10} \, \text{s}

Step 4: Use the Given Ratio of Refractive Indices

We are given that the ratio of the refractive indices \( \frac{\mu_A}{\mu_B} = \frac{1}{2} \). In terms of velocities, since:

• \mu_A = \frac{c}{v_A}
• \mu_B = \frac{c}{v_B}

We can substitute the given ratio:

\frac{\frac{c}{v_A}}{\frac{c}{v_B}} = \frac{v_B}{v_A} = \frac{1}{2} \Rightarrow v_B = 2v_A

Step 5: Substitute and Solve for \( d \)

Substituting \( v_B = 2v_A \) in the time difference equation:

d \left( \frac{1}{2v_A} - \frac{1}{v_A} \right) = 5 \times 10^{-10} \, \text{s}

Simplifying inside the brackets:

d \left( \frac{1}{2v_A} - \frac{2}{2v_A} \right) = d \left( -\frac{1}{2v_A} \right) = 5 \times 10^{-10} \, \text{s}

\Rightarrow -\frac{d}{2v_A} = 5 \times 10^{-10} \, \text{s} \Rightarrow d = -2v_A \times 5 \times 10^{-10} \, \text{s} = 5 \times 10^{-10} v_A \, \text{m}

Hence, the thickness of the material is 5 \times 10^{-10} \, v_A \, \text{m} .