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JEE Main 2022 · 24 Jun · Shift 2, Physics question 57

A ray of light is incident at an angle of incidence 60° on the glass slab of refractive index 3\sqrt3. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 434\sqrt3 cm. The thickness of the glass slab is ______ cm.

Answer and worked solution

Official answer: 12

Refraction angle. sin⁡r=sin⁡60°3=12\sin r=\dfrac{\sin60°}{\sqrt3}=\dfrac12, so r=30°r=30°.

Lateral shift. d=tsin⁡(i−r)cos⁡r=tsin⁡30°cos⁡30°=t3d=\dfrac{t\sin(i-r)}{\cos r}=\dfrac{t\sin30°}{\cos30°}=\dfrac t{\sqrt3}.

Thickness. t3=43\dfrac t{\sqrt3}=4\sqrt3 gives t=12t=12 cm.

Answer: 12

Topics:Refraction at plane surfaces
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