Question

Find the time taken by light to cover a distance of 9 mm in water. Take ${\mu }_{water}$ = $\frac{4}{3}$

• A. 0.04 ns
• B. 0.4 ns
• C. 4 ns
• D. 400 ns

Answer 1

The correct option is A

0.04 ns

${\mu }_1$​ (Velocity of light in medium 1) = ${\mu }_2$​ (Velocity of light in medium 2)

${\mu }_{air}$ $\times$ $v_{air}$ = ${\mu }_{water}$ $\times$ $v_{water}$

1 $\times$ (3$\times$${10}^8)=\frac{4}{3}$ $\times$ $v_{water}$

$v_{water}$ = $\frac{9}{4}$ $\times$ ${10}^8$ $\frac{m}{s}$

We know, v = $\frac{d}{t}$; t = $\frac{d}{v}$ = $\frac{9\times {10}^{-3}}{\frac{9}{4}\times {10}^8}$ = 4 $\times$ ${10}^{-11}$ sec

t = 0.04 ns