Question

Question #7abf9

Answer 1

`"Critical angle"=pi/6=30^circ`

Explanation:

The equation for critical angle is:
`sinC=n_1/n_2`, where:

• `C` = critical angle
• `n_1` = boundary medium
• `n_2` = initial medium

In this case `n_1` is the vacuum as it is the medium acting as a boundary, while `n_2` is the other medium, as the light is travelling through that medium.

`n_1=1`

`sinC=1/2`

`C=sin^(-1)(1/2)=pi/6=30^circ`