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  4. A ray of light enters a rectangular glass slab of refractive index √3 at an angle of incidence 60°.It travels a distance 5 cm inside and emerges out of slab. The perpendicular distance between the incidence and the emergent ray is

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Q. A ray of light enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{\circ}$.It travels a distance $5 \,cm$ inside and emerges out of slab. The perpendicular distance between the incidence and the emergent ray is

Solution:

$\sin 60^{\circ}=n \sin r $
$\Rightarrow r=30^{\circ}$
$x=d \sin (i-r)=5 \times \sin 30^{\circ}=5 / 2 \,cm$