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A ray of light is incident from a rarer medium $(μ = 1.5)$ into a denser medium. If the angle of incidence and refraction are respectively $60^{o}$ and $45^{o},$ then calculate the refractive index of the denser medium.

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$3 \sqrt{3} / 2 \sqrt{2}$

$\sqrt{3}$

$1$

$\frac{3}{2}$

answer is D.

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Detailed Solution

$μ_{r a r e r} = 1.5, i = 60^{0}, r = 45^{0}$

$μ_{d e n s e r} = ?$

$μ_{r a r e r} \sin i = μ_{d e n s e r} \sin r$

$\Rightarrow 1.5 \times \sin 60^{0} = μ_{d e n s e r} \sin 45^{0}$

$μ_{d e n s e r} = \frac{1.5 \times 60^{0}}{\sin 45^{0}}$

$μ_{d e n s e r} = \frac{1.5 \times (\sqrt{3} / 2)}{(1 / \sqrt{2})} =$ $\frac{1.5 \times \sqrt{3} \times \sqrt{2}}{2} = 1.84$

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