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A ray of light is incident at an angle of incidence 60^{\circ} on the glass slab of refractive index \sqrt{3}. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 4\sqrt{3} cm. The thickness of the glass slab is __________cm.

Option: 1

12


Option: 2

--


Option: 3

--


Option: 4

--


Answers (1)

best_answer

\begin{array}{l|l} i=60^{\circ} & \text { By snell's law } \\ & \mu_{1} \sin i=\mu _{2}sin r \\ \mu_{2}=\sqrt{3} & 1 \times \sin 60^{\circ}=\sqrt{3}sinr \\ & r=30^{\circ} \end{array}

$$ \begin{aligned} \Delta L &=\text { Lateral shift }=4 \sqrt{3} \mathrm{~cm} \\ 4 \sqrt{3} &=\frac{t \sin (i-r)}{\cos r} \\ 4 \sqrt{3} &=\frac{t \sin 30^{\circ}}{\cos 30^{\circ}} \\ t &=12 \mathrm{~cm} \end{aligned}


The thickness of the glass slab is 12 \mathrm{~cm}

Posted by

Ritika Kankaria

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