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JEE Main 2022 (24 Jun, Shift 2) · Refraction, Glass Slab & Apparent Depth

Refraction, Glass Slab & Apparent Depth question from JEE Main 2022 (24 Jun, Shift 2)

Q50JEE Main 2022 · 24 JUN · SHIFT 2Medium

Refraction, Glass Slab & Apparent Depth

A ray of light is incident at an angle of incidence 60∘^\circ on the glass slab of refractive index 3\sqrt{3}. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 43\sqrt{3} cm. The thickness of the glass slab is __________ cm.

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

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Answer: 12

60°30°dtμ = √3
  1. Snell's law: sin⁡r=sin⁡60∘3=12⇒r=30∘\sin r = \dfrac{\sin 60^\circ}{\sqrt{3}} = \dfrac{1}{2} \Rightarrow r = 30^\circ.
  2. Lateral shift d=tsin⁡(i−r)cos⁡r=tsin⁡30∘cos⁡30∘=t3d = \dfrac{t\sin(i - r)}{\cos r} = \dfrac{t\sin 30^\circ}{\cos 30^\circ} = \dfrac{t}{\sqrt{3}}.
  3. 43=t3⇒t=124\sqrt{3} = \dfrac{t}{\sqrt{3}} \Rightarrow t = 12.
  4. Answer: 12 cm.
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