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  • A ray of light enters a rectangular slab of refractive index  at an angle of incidence 60o.

A ray of light enters a rectangular slab of refractive index \sqrt{3} at an angle of incidence 60o. It travels a distance of 5cm inside the slab and emerges out of the slab. The perpendicular distance between incident and emergent rays is:

Option: 1

5\sqrt{3}cm

 

 

 


Option: 2

\frac{5}{2}cm


Option: 3

5\sqrt{}\frac{5}{2}cm


Option: 4

5cm


Answers (1)

best_answer

As we learn

Lateral displacement -

\Delta t= \frac{t}{\cos r}\sin \left ( i-r \right )

- wherein

t= thickness of slab

i= angle of incidence

r = angle of refraction

 

 \frac{\sin 60^{0}}{\sin r_{1}}=\sqrt{3}

or \sin r_{1}=\frac{\sin 60^{0}}{\sqrt{3}}=\frac{1}{2}

orr_{1}=30^{0}

now \sin (60^{0}-30^{0})=\frac{d}{5}

                 

\Rightarrow \frac{d}{5}=\frac{1}{2}

\Rightarrow d= \frac{5}{2}cm

 

Posted by

seema garhwal

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