Q55 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
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 $\_\_\_\_$.
- A$20^{\circ}$
- Correct answerB$15^{\circ}$
- C$40^{\circ}$
- D$30^{\circ}$
Solution
Given an equilateral prism, the angle of the prism is $A = 60^{\circ}$.
The refractive index of the prism is $\mu = \sqrt{2}$.
The emergent ray grazes the second surface, which means the angle of emergence is $e = 90^{\circ}$.
Applying Snell's law at the second surface: $\mu \sin(r_2) = 1 \cdot \sin(e)$.
$\sqrt{2} \sin(r_2) = \sin(90^{\circ}) = 1$.
$\sin(r_2) = \frac{1}{\sqrt{2}}$, which gives $r_2 = 45^{\circ}$.
For a prism, we know the relation $r_1 + r_2 = A$.
Substituting the values: $r_1 + 45^{\circ} = 60^{\circ}$.
$r_1 = 60^{\circ} - 45^{\circ} = 15^{\circ}$.
The angle of refraction at the incident surface is $r_1 = 15^{\circ}$.
The refractive index of the prism is $\mu = \sqrt{2}$.
The emergent ray grazes the second surface, which means the angle of emergence is $e = 90^{\circ}$.
Applying Snell's law at the second surface: $\mu \sin(r_2) = 1 \cdot \sin(e)$.
$\sqrt{2} \sin(r_2) = \sin(90^{\circ}) = 1$.
$\sin(r_2) = \frac{1}{\sqrt{2}}$, which gives $r_2 = 45^{\circ}$.
For a prism, we know the relation $r_1 + r_2 = A$.
Substituting the values: $r_1 + 45^{\circ} = 60^{\circ}$.
$r_1 = 60^{\circ} - 45^{\circ} = 15^{\circ}$.
The angle of refraction at the incident surface is $r_1 = 15^{\circ}$.
Hence, the correct answer is option: 'B'.