Question

Light falls at normal incidence on one face of a glass prism of refractive index 1.5, what will be the angle of emergence when the angle of the prism is (a) `30^(@)` (b) `50^(@)` ?

Answer 1

The incident ray is normal to the face AB of the prism shown in the figure. So, the angle of incidence, `i_(1) = 0` making `r_(1) = 0`
When light undergoes refraction through a prism, `r_(1) + r_(2) = A`
(a) `0 + r_(2) = 30^(@)`
`r_(2) = 30^(@)`
`sin i_(2) = mu sin r_(2) = 1.5 sin 30^(@) = 1.5 (0.5) = 0.75`
(b) `A = 50^(@)`
Substituting in `r_(1) + r_(2) = A`
`0 + r_(2) = 50^(@), r_(2) = 50^(@)`
`sin i_(2) = mu sin r_(2) = 1.5 sin 50^(@) = 1.5 (0.7660) = 1.149`
No value of `i_(2)` can satisfy this relation. So, there is not refraction at the second surface i.e., there is no emergent ray from the face AC of the prism. The light is totally reflected as `r_(2)` is greater than the critical angle.