1.

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

Answer»

`2.4xx10^8 m s^(-1)`
`3.0 xx 10^8 m s^(-1)`
`1.2xx10^8 m s^(-1)`
`1.8 xx 10^8 m s^(-1)`

Solution :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 fromthe 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)`


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