Velocity of light in a liquid is `1.5 xx 10^(8) m//s` and in air, it is `3 xx 10^(8)m//s`. If a ray of light passes from this liquid to air, calculate the value of critical angle.

Answer

`30^@`

Explanation

Here, `v = 1.5 xx 10^8 m//s`
`c = 3 xx 10^8 m//s`
`mu = ( c)/(v) = (3 xx 10^8)/(1.5 xx 10^8) = 2`
If `C` is crictical angle, then
`sin C = (1)/(mu) = (1)/(2)`
`C = 30^@`.