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Physics Question on Refraction of Light

The time required for the light to pass through a glass slab (refractive index = 1.51.5) of thickness 4 mm4 \,mm is (c=3×108 ms−1c = 3 \times 10^8\, ms^{-1}, speed of light in free space).

A

10−11sec10^{-11} sec

B

2×10−11 sec2\times10^{-11}\, sec

C

2×10+11 sec2\times10^{+11} \,sec

D

2×10−5 sec2\times10^{-5} \,sec

Answer

2×10−11 sec2\times10^{-11}\, sec

Explanation

Solution

We know, naca=ngcgn_{a} c_{a}=n_{g} c_{g} ngna=cacg\frac{n_{g}}{n_{a}}=\frac{c_{a}}{c_{g}} 32=3×108cg\frac{3}{2}=\frac{3 \times 10^{8}}{c_{g}} cg=2×108c_{g}=2 \times 10^{8} We have,Time = Distance  Speed =\frac{\text { Distance }}{\text { Speed }} t=4×10−32×108t=\frac{4 \times 10^{-3}}{2 \times 10^{8}} Or t=2×10−11st=2 \times 10^{-11} s