Ray optics

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A light beam is incident on a denser medium whose refractive index is 1.414 at an angle of incidence 45°. Find the ratio of width of refracted beam in a medium to the width of the incident beam in air.

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Solution

$\begin{array}{l} μ = 1.414, i = 45^{0} \\ \frac{\sin i}{\sin r} = μ \\ \frac{1}{\sqrt{2}} = 1.414 \sin r \\ \sin r = \frac{1}{2} \Rightarrow r = 30^{0} \end{array}$
$\sin 45^{0} = \frac{A B}{A C}, \cos 30^{0} = \frac{C D}{A C}$
Width in rarer medium $A B = \frac{A C}{\sqrt{2}}$
Width in denser medium $C D = \frac{\sqrt{3} A C}{2}$
$\frac{C D}{A B} = \frac{\sqrt{3} / 2}{1 / \sqrt{2}} = \sqrt{\frac{3}{2}}$

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Ray optics

Remember concepts with our Masterclasses.

A light beam is incident on a denser medium whose refractive index is 1.414 at an angle of incidence 45°. Find the ratio of width of refracted beam in a medium to the width of the incident beam in air.

μ=1.414, i=450sinisinr=μ12 = 1.414 sinrsin r=12 ⇒ r = 300sin450=ABAC, cos300=CDACWidth in rarer medium AB=AC2 Width in denser medium CD=3AC2CDAB=3/21/2 = 32

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