First slide

Refraction of light

Question

A ray incident on a droplet of water at an angle of incidence $i$ undergoes two reflections (not total) and emerges. If the deviation suffered by the ray within the drop is minimum and the refractive index of the droplet be $μ$ , then $\cos i = \sqrt{\frac{μ^{2} - 1}{x}}$ . Find the value of $x$ .

Difficult

Solution

$\begin{array}{l} δ_{Total} = δ_{Refraction} + 2 δ_{Reflection} + δ_{Refraction} \\ \Rightarrow δ = (i - r) + 2 (180 ° - 2 r) + (i - r) \\ \Rightarrow δ = 360 ° + 2 i - 6 r \\ \Rightarrow δ = 360 ° + 2 i - 6 \sin^{- 1} (\frac{\sin i}{μ}) \\ For deviation to be minimum, \frac{d δ}{d i} = 0 \\ \Rightarrow 2 - \frac{6}{\sqrt{1 - {(\frac{\sin i}{μ})}^{2}}} \frac{\cos i}{μ} = 0 \\ \Rightarrow \cos i = \sqrt{\frac{μ^{2} - 1}{8}} \\ ∴ x = 8 \end{array}$

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