Optics Ques 49
- A right angled prism is to be made by selecting a proper material and the angles $A$ and $B(B \leq A)$, as shown in figure. It is desired that a ray of light incident on the face $A B$ emerges parallel to the incident direction after two internal reflections.
(1987, 7M)
(a) What should be the minimum refractive index $n$ for this to be possible?
(b) For $n=5 / 3$ is it possible to achieve this with the angle $B$ equal to $30$ degrees?
Show Answer
Answer:
Correct Answer: 49.(a) $\sqrt{2}$
(b) No
Solution:
Formula:
- (a) At $P$, angle of incidence $i _A=A$ and at $Q$, angle of incidence $i _B=B$
If TIR satisfies for the smaller angle of incidence than for larger angle of incidence is automatically satisfied.
$ B \leq A \quad \therefore \quad i _B \leq i _A $
Maximum value of $B$ can be $45^{\circ}$. Therefore, if condition of TIR is satisfied, then condition of TIR will be satisfied for all value of $i _A$ and $i _B$
$\text { Thus, } 45^{\circ} \geq \theta _c $
$\text { or } \sin 45^{\circ} \geq \sin \theta _c $
$\text { or } \frac{1}{\sqrt{2}} \geq \frac{1}{\mu} \text { or } \mu \geq \sqrt{2}$
$\therefore \quad$ Minimum value of $\mu$ or $n$ is $\sqrt{2}$.
(b) For $n=\frac{5}{3}, \sin \theta _c=\frac{1}{n}=\sin ^{-1} (\frac{3}{5}) \approx 37^{\circ}$
If $\quad B=30^{\circ}$, then $i _B=30^{\circ}$
then $A=60^{\circ}$ or $i _A=60^{\circ}$
$i _A>\theta _c$ but $i _B<\theta _c$
i.e. TIR will take place at $A$ but not at $B$.
or we write : $\sin i _B<\sin \theta _c<\sin i _A$
or $ \quad \sin 30^{\circ}<\frac{3}{5}<\sin 60^{\circ} $
or $\quad 0.5 < 0. 6 <0 .86 $
BETA
