Ray Optics and Optical Instruments Part 2

Given, the base of a prism is equilateral triangle of side 10 cm as shown in figure. At minimum deviation, Angle of incident = Angle of prism i.e. i = A = 60° Angle of refraction, $r = \frac{A}{2} = \frac{60^{\circ}}{2} = 30^{\circ}$ Let µ be the refractive index. Then, by Snell’s law of refraction,   $\mu = \frac{\sin i}{\sin r} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$ Now, in ∆ABP, AP = AB cos 30° $= \frac{10 \sqrt{3}}{2} = 5 \sqrt{3} \text{cm}$ $= 5 \sqrt{3} \times 10^{-2} \text{m}$ i.e. optical distance travelled by light along AP, $d = \mu \times AP = \sqrt{3} \times 5 \sqrt{3} \times 10^{-2} = 15 \times 10^{-2} \text{m}$ Now, the time taken by light $(c = 3 \times 10^{8} \text{m/s})$ to travel from P to A, $t = \frac{d}{C} = \frac{15 \times 10^{-2}}{3 \times 10^{8}} \text{s} = 5 \times 10^{-10} \text{s}$ Thus, correct answer is 5.