JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2021
- Hard
- Numerical
Cross-section view of a prism is the equilateral triangle \({ABC}\) in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from midpoint of \({BC}\) to \({A}\) is \(.....\times 10^{-10}\, {s}\). \((\) Given, speed of light in vacuum \(=3 \times 10^{8} \,{m} / {s}\) and \(\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)\)

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Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
\({i}={A}=60^{\circ}\)
\(={\delta}_{\min }=2 {i}-{A}\)
\(=2 \times 60^{\circ}-60^{\circ}=60^{\circ}\)
\(\mu=\frac{\sin ^{-1}\left(\frac{\delta_{\min }+{A}}{2}\right)}{\sin ^{-1}\left(\frac{{A}}{2}\right)}\)
\(=\sqrt{3}\)
\({V}_{\text { prism }}=\frac{3 \times 10^{8}}{\sqrt{3}}\)
\({AP}=10 \times 10^{-2} \times \frac{\sqrt{3}}{2}\)
\(\text { time }=\frac{5 \times 10^{-2}}{3 \times 10^{8}} \times \sqrt{3} \times \sqrt{3}\)
\(=5 \times 10^{-10}\, {sec}\)
\(\text { Ans }=5\)
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