A light ray passes from air to denser medium of certain thickness and emerges on the other side. If the emergent ray is parallel to the incident ray, the distance traveled by the ray of light in the denser medium is 6 cm, and the angle of incidence and refraction are $60^{0}$ and $30^{0}$ respectively, find the lateral displacement of the light ray.

9 cm

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2 cm

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3 cm

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6 cm

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Solution

The correct option is C 3 cm

$t = 6 cos 30^{0} c m$ [as per question 6cm is distance travelled by light ray in medium]
$t = 3 \sqrt{3} c m$
Hence
Lateral shift $δ = \frac{3 \sqrt{3}}{cos 30^{0}} sin 30^{0} = \frac{3 \sqrt{3}}{\frac{\sqrt{3}}{2}} / \frac{1}{2}$
$δ = 3 c m .$

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The correct option is C 3 cmt=6cos300cm[as per question 6cm is distance travelled by light ray in medium]t=3√3cmHenceLateral shift δ=3√3cos300sin300=3√3√32/12δ=3cm.

The correct option is C 3 cm

$t = 6 cos 30^{0} c m$ [as per question 6cm is distance travelled by light ray in medium]
$t = 3 \sqrt{3} c m$
Hence
Lateral shift $δ = \frac{3 \sqrt{3}}{cos 30^{0}} sin 30^{0} = \frac{3 \sqrt{3}}{\frac{\sqrt{3}}{2}} / \frac{1}{2}$
$δ = 3 c m .$