Time taken by light to travel in two different materials A and B of refractive indices μA and μB 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 vA and vB are velocities of light in A and B materials respectively).

μAμB=c/VAc/VB=VBVA=12Let the thickness is ddvB−dvA=5×10−10d=5×10−10×vAvBvA−vB As vA=2vB⇒d=5×10−10×2vB Or d=5×10−10×vA

Time taken by light to travel in two different materials A and B of refractive indices μA and μB 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 vA and vB are velocities of light in A and B materials respectively).

μAμB=c/VAc/VB=VBVA=12Let the thickness is ddvB−dvA=5×10−10d=5×10−10×vAvBvA−vB As vA=2vB⇒d=5×10−10×2vB Or d=5×10−10×vA

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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 $t_{2} - t_{1} = 5 \times 10^{- 10} s$ and the ratio of $μ_{A} to μ_{B}$ 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).

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$5 \times 10^{- 10} v_{B} m$

$5 \times 10^{- 10} m$

$5 \times 10^{- 10} v_{A} m$

$1.5 \times 10^{- 10} m$

answer is A.

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Detailed Solution

$\frac{μ_{A}}{μ_{B}} = \frac{c / V_{A}}{c / V_{B}} = \frac{V_{B}}{V_{A}} = \frac{1}{2}$

Let the thickness is d

$\begin{array}{l} \frac{d}{v_{B}} - \frac{d}{v_{A}} = 5 \times 10^{- 10} \\ d = \frac{5 \times 10^{- 10} \times v_{A} v_{B}}{v_{A} - v_{B}} \\ As v_{A} = 2 v_{B} \Rightarrow d = 5 \times 10^{- 10} \times 2 v_{B} \\ Or d = 5 \times 10^{- 10} \times v_{A} \end{array}$