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JEE Main 2024 (4 Apr, Shift 2) · Total Internal Reflection

Total Internal Reflection question from JEE Main 2024 (4 Apr, Shift 2)

Q18JEE Main 2024 · 4 APR · SHIFT 2Medium

Total Internal Reflection

A light ray is incident on a glass slab of thickness 43 cm4\sqrt{3}\text{ cm} and refractive index 2\sqrt{2} The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ______ cm\text{cm}.

(Given sin⁡15∘=0.25\sin 15^\circ = 0.25)

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

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Answer: 2

  1. i=Ci = C: sin⁡C=12⇒i=45∘\sin C = \frac{1}{\sqrt{2}} \Rightarrow i = 45^\circ.
  2. sin⁡r=sin⁡45∘2=12⇒r=30∘\sin r = \frac{\sin45^\circ}{\sqrt{2}} = \frac{1}{2} \Rightarrow r = 30^\circ.
  3. d=tsin⁡(i−r)cos⁡r=43×0.253/2=2d = \frac{t\sin(i-r)}{\cos r} = \frac{4\sqrt{3}\times0.25}{\sqrt{3}/2} = 2 cm.
  4. So the lateral displacement is 2 cm.
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