Q65 / 75 · JEE Main 2026 (05 April Shift 1) · Physics
A ray of light passing through an equilateral prism is having velocity $2.12 \times 10^8$ m/s in the prism material, then the minimum angle of deviation is _________ degrees.
- A$45$
- Correct answerB$30$
- C$28$
- D$58$
Solution
Velocity of light in vacuum, $c = 3 \times 10^8$ m/s
Velocity of light in the prism, $v = 2.12 \times 10^8$ m/s
Refractive index of the prism material, $\mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}$
For an equilateral prism, the angle of the prism $A = 60^\circ$.
Using the prism formula:
$\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$
$\sqrt{2} = \dfrac{\sin\left(\dfrac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}$
$\sqrt{2} = \dfrac{\sin\left(30^\circ + \dfrac{\delta_m}{2}\right)}{\dfrac{1}{2}}$
$\sin\left(30^\circ + \dfrac{\delta_m}{2}\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$
$30^\circ + \dfrac{\delta_m}{2} = 45^\circ$
$\dfrac{\delta_m}{2} = 15^\circ$
$\delta_m = 30^\circ$
Answer: $30$
Velocity of light in the prism, $v = 2.12 \times 10^8$ m/s
Refractive index of the prism material, $\mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}$
For an equilateral prism, the angle of the prism $A = 60^\circ$.
Using the prism formula:
$\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$
$\sqrt{2} = \dfrac{\sin\left(\dfrac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}$
$\sqrt{2} = \dfrac{\sin\left(30^\circ + \dfrac{\delta_m}{2}\right)}{\dfrac{1}{2}}$
$\sin\left(30^\circ + \dfrac{\delta_m}{2}\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$
$30^\circ + \dfrac{\delta_m}{2} = 45^\circ$
$\dfrac{\delta_m}{2} = 15^\circ$
$\delta_m = 30^\circ$
Answer: $30$
Hence, the correct answer is option: 'B'.