Question 90: Ray Optics

Question

Consider an equilateral prism (refractive index $\sqrt{2}$ ). A ray of light is incident on its one surface at a certain angle $i$. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to $\_\_\_\_$

Accepted Answer

The given prism is an equilateral, that means the angle of the prism (A) is $60^{\circ}$.
Refractive index of the prism is $\mu=\sqrt{2}$
The ray grazes the second surface. This means the angle of emergence (e) is $90^{\circ}$.
The angle of incidence at the first surface of prism is i .
Let the angle of refraction at first surface is $\mathrm{r}_1$.
The ray travels straight through the prism and hits the second face AC.
Let the angle of incidence at the second surface be $\mathrm{r}_2$.
The ray bends away from the normal as it exits into the air. Let this angle be e (angle of emergence).
The sum of angles $r_1$ and $r_2$ is equal to the angle of prism.
$$ \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 $$
Using Snell's law at the emergent surface (Face AC),
$$ \sqrt{2} \cdot \sin \left(r_2\right)=1 \cdot \sin \left(90^{\circ}\right) $$
$$\Rightarrow $$ $\sqrt{2} \cdot \sin \left(r_2\right)=1$
$$\Rightarrow $$ $$ \sin \left(r_2\right)=\frac{1}{\sqrt{2}} \Rightarrow r_2=45^{\circ} $$
Using $\mathrm{A}=\mathrm{r}_1+\mathrm{r}_2$
$$ 60^{\circ}=\mathrm{r}_1+45^{\circ} $$
$$\Rightarrow $$ $r_1=60^{\circ}-45^{\circ}$
$$\Rightarrow $$ $$ r_1=15^{\circ} $$
So, the angle of refraction at the incident surface is $15^{\circ}$.
Hence, the correct option is (D).

Answer

$30^{\circ}$

Answer

$20^{\circ}$

Answer

$40^{\circ}$

Answer

$15^{\circ}$

Ray Optics

Consider an equilateral prism (refractive index \sqrt{2} ). A ray of light is incident on its one surface at a certain angle i. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to \_\_\_\_

Choose the correct answer: