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JEE Main 2023 (25 Jan, Shift 1) · Total Internal Reflection

Total Internal Reflection question from JEE Main 2023 (25 Jan, Shift 1)

Q22JEE Main 2023 · 25 JAN · SHIFT 1Medium

Total Internal Reflection

A ray of light is incident from air on a glass plate having thickness 3\sqrt{3} cm and refractive index 2\sqrt{2}. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ____________ ×\times 10 −2-2 cm. (given sin⁡15∘=0.26\sin 15^\circ = 0.26)

Numerical answer type. Work it out, then open the answer.

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Answer: 52

  1. Critical angle for glass-air: sin⁡C=12⇒i=C=45∘\sin C = \frac{1}{\sqrt{2}} \Rightarrow i = C = 45^\circ.
  2. Air to glass: sin⁡r=sin⁡45∘2=12⇒r=30∘\sin r = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1}{2} \Rightarrow r = 30^\circ.
  3. Lateral shift d=tsin⁡(i−r)cos⁡r=3×sin⁡15∘3/2=2×0.26=0.52d = \frac{t\sin(i-r)}{\cos r} = \frac{\sqrt{3}\times\sin15^\circ}{\sqrt{3}/2} = 2\times0.26 = 0.52 cm.
  4. So the answer is 52×10−252\times10^{-2} cm, i.e. 52.
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