Optics 4 Question 14
14. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face $A B$ is totally reflected to reach the face $B C$ if
(1981, 3M)
(a) $\sin \theta>8 / 9$
(b) $2 / 3<\sin \theta<8 / 9$
(c) $\sin \theta<2 / 3$
(d) None of these
Objective Question II (One or more correct option)
Show Answer
Solution:
- Let $\theta _c$ be the critical angle at face $B C$, then
Angle of incidence at face $B C$ is $i=\theta$
Total internal reflection (TIR) will take place on this surface if, $i>\theta _c$ or $\theta>\theta _c$ or $\sin \theta>\sin \theta _c$ or $\sin \theta>\frac{8}{9}$
15 The minimum deviation produced by a prism
$$ \begin{aligned} \delta _m & =2 i-A=A \\ \therefore \quad i _1 & =i _2=A \text { and } \\ r _1 & =r _2=A / 2 \\ \therefore \quad r _1 & =i _1 / 2 \end{aligned} $$
Now, using Snell’s law
$$ \sin A=\mu \sin A / 2 $$
$\Rightarrow \quad \mu=2 \cos (A / 2)$
For this prism when the emergent ray at the second surface is tangential to the surface
$$ \begin{aligned} & \quad i _2=\pi / 2 \Rightarrow r _2=\theta _c \Rightarrow r _1=A-\theta _c \\ & \text { so, } \sin i _1=\mu \sin \left(A-\theta _c\right) \\ & \text { so, } \quad i _1=\sin ^{-1} \sin A \sqrt{4 \cos ^{2} \frac{A}{2}-1}-\cos A \end{aligned} $$
For minimum deviation through isosceles prism, the ray inside the prism is parallel to the base of the prism if $\angle B=\angle C$.
But it is not necessarily parallel to the base if,
$$ \angle A=\angle B \text { or } \angle A=\angle C $$
