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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2019 · Paper 2 · Physics · Ray Optics

A monochromatic light is incident from air on a refracting surface...

A monochromatic light is incident from air on a refracting surface of a prism of angle 75o{75}^{o} and refractive index n0=3.{n}_{0} = \sqrt{3} . The other refracting surface of a prism is coated by a thin film of material of refractive index nn as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of θ≤60o.\theta \le {60}^{o} . The value of n2{n}^{2} is_______.
A monochromatic light is incident from air on a refracting surface of a prism of angle 75 ^ o and refractive index n _

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Step-by-step solution

A monochromatic light is incident from air on a refracting surface of a prism of angle 75 ^ o and refractive index n _
For T.I.R at coating
sinc=n3sinc = \dfrac{n}{\sqrt{3}} …(i) \left[sin C = \dfrac{{n}_{2}}{{n}_{1}}\right]<br/>ApplingSnell′slawatfirstsurface<br/><br/>Appling Snell's law at first surface<br/>sin \theta = \sqrt{3}\sin \left(75 - c\right)<br/>forlimitingconduction,at<br/>for limiting conduction, at\theta = {60}^{o}<br/><br/>\sin 60 = \sqrt{3}\sin \left(75 - c\right)<br/><br/>\dfrac{\sqrt{3}}{2} = \sqrt{3}\sin \left(75 - c\right)<br/><br/>\begin{cases}\dfrac{1}{2} = \sin \left(75 - c\right) & ⇒ & \sin 30 = \sin \left(75 - c\right) \\ ⇒ 30 = 75 - c & ⇒ & c = {45}^{o} \\ Using\left(i\right)\dfrac{n}{\sqrt{3}} = \dfrac{1}{\sqrt{2}} & ⇒ & {n}^{2} = \dfrac{3}{2} = 1.50\end{cases}
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