Question 83: Ray Optics

Question

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\frac{n}{2}$. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is $\_\_\_\_$ .

Accepted Answer

Light is traveling from the prism material (denser medium) to the coating material (rarer medium).
Refractive index of Prism $\mathrm{n}_1=\mathrm{n}$
Refractive index of Coating $\mathrm{n}_2=\frac{\mathrm{n}}{2}$
According to question at the exit surface, the light exactly meets the condition of critical angle.
The formula for the critical angle is :
$$ \sin 	heta_{\mathrm{C}}=\frac{\mathrm{n}_{	ext {rare }}}{\mathrm{n}_{	ext {denser }}} $$
$$\Rightarrow $$ $\sin 	heta_C=\frac{n / 2}{n}=\frac{1}{2}$
$$ 	heta_C=\sin ^{-1}(1 / 2)=30^{\circ} $$
So, the angle of incidence at the second face $\left(\mathrm{r}_2ight)$ is equal to the critical angle :
$$ r_2=30^{\circ} $$
Applying the condition for minimum deviation, the light ray passes symmetrically through the prism.
$$ \mathrm{r}_1=\mathrm{r}_2=30^{\circ} $$
The light ray inside the prism is parallel to the base (for an equilateral prism).
For any prism, the relationship between the Prism Angle A and the internal angles is:
$$ \begin{gathered} \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 \ \mathrm{~A}=30^{\circ}+30^{\circ} \ \mathrm{A}=60^{\circ} \end{gathered} $$
Hence, the correct option is (A).

Answer

$60^{\circ}$

Answer

$30^{\circ}$

Answer

$15^{\circ}$

Answer

$45^{\circ}$

Ray Optics

The exit surface of a prism with refractive index n is coated with a material having refractive index \frac{n}{2}. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is \_\_\_\_ .

Choose the correct answer: