First slide

Total internal reflection - Ray optics and optical instruments

Question

In the figure shown, for an angle of incidence 45° at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face

Difficult

A

$\frac{\sqrt{2} + 1}{2}$
B

$\sqrt{\frac{3}{2}}$
C

$\sqrt{\frac{1}{2}}$
D

$\sqrt{2} + 1$

Solution

At point A, by Snell’s law
$μ = \frac{\sin 45}{sinr} \Rightarrow sinr = \frac{1}{μ \sqrt{2}}$ ----------(i)
$From figure, i_{1} = 90 - r ∴ \sin (90^{\circ} - r) = \frac{1}{μ} \Rightarrow cosr = \frac{1}{μ} - - - - - - - (ii) Now cosr = \sqrt{1 - \sin^{2} r} = \sqrt{1 - \frac{1}{2 μ^{2}}} = \sqrt{\frac{2 μ^{2} - 1}{2 μ^{2}}} - - - - (iii) From equation (ii) and (iii) \frac{1}{μ} = \sqrt{\frac{2 μ^{2} - 1}{2 μ^{2}}} Squaring both side and then solving we get μ = \sqrt{\frac{3}{2}}$

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