Ray Optics

JEE Main level, numerical answer, medium

255 of 295

Question 255 of 295

Cross-section view of a prism is the equilateral triangle ABCABC shown 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 PP (midpoint of BCBC) to AA is ___________ ×10−10s.\times {10}^{-10}s. (Given, speed of light in vacuum =3×108ms−1=3\times {10}^{8}m{s}^{-1} and cos⁡30∘=32\cos 30^{\circ}=\dfrac{\sqrt{3}}{2})

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Answer: 5

δmin=2i−A{\delta }_{min}=2i-A
A=2i−AA=2i-A
A=iA=i
μ=sin⁡(δmin+A2)sin⁡(A2)=sin⁡(A)sin⁡(A2)=sin⁡60∘sin⁡30∘\mu =\dfrac{\sin \left(\dfrac{{\delta }_{min}+A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)}=\dfrac{\sin (A)}{\sin \left(\dfrac{A}{2}\right)}=\dfrac{\sin 60^{\circ}}{\sin 30^{\circ}}
μ=3\mu =\sqrt{3}
v=cμ=3×1083ms−1v=\dfrac{c}{\mu }=\dfrac{3\times {10}^{8}}{\sqrt{3}}m{s}^{-1}
t=APv=53×10−2(3×1083)=5×10−10st=\dfrac{AP}{v}=\dfrac{5\sqrt{3}\times {10}^{-2}}{\left(\dfrac{3\times {10}^{8}}{\sqrt{3}}\right)}=5\times {10}^{-10}\text{s}
So, x=5x=5

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