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JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments

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 \(\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)\)

  1. A \(0.005\)
  2. B \(500\)
  3. C \(5.173\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(5\)

Step-by-step Solution

Detailed explanation

\({i}={A}=60^{\circ}\) \(={\delta}_{\min }=2 {i}-{A}\) \(=2 \times 60^{\circ}-60^{\circ}=60^{\circ}\) \(\mu=\frac{\sin ^{-1}\left(\frac{\delta_{\min }+{A}}{2}\right)}{\sin ^{-1}\left(\frac{{A}}{2}\right)}\) \(=\sqrt{3}\) \({V}_{\text { prism }}=\frac{3 \times 10^{8}}{\sqrt{3}}\)…
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