A small coin is resting on the bottom of a beaker filled with liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?

Options

A. `2.4xx10^8 m s^(-1)`
B. `3.0 xx 10^8 m s^(-1)`
C. `1.2xx10^8 m s^(-1)`
D. `1.8 xx 10^8 m s^(-1)`

Answer

D

Explanation

As sin `i_c=3/sqrt(3^2+4^2)=3/5 therefore mu=1/(sin i_c)=5/3`
(as `i_c` is the angle which the ray from the coin makes with 4 cm side)
As refractive index , `mu=c/v_l`
`therefore v_l=c/mu=(3xx10^8 m s^(-1))/((5//3))=1.8xx10^8 m s^(-1)`