For a given incident ray as shown in Figure ,  the condition of total internal reflection of ray will be …

Q: For a given incident ray as shown in Figure ,  the condition of total internal reflection of ray will be satisfied if the refractive index of the block will be

Numerical

$\displaystyle (a) \frac{\sqrt{3}+1}{2}$

$\displaystyle (b) \frac{\sqrt{2}+1}{2}$

$\displaystyle (c) \sqrt{\frac{3}{2}}$

$\displaystyle (d) \sqrt{\frac{7}{6}}$

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Ans: (c)

Sol: Here , r + C = 90°

⇒ r = 90° – C

Numerical

Applying Snell’s Law at B ,

$\displaystyle \mu = \frac{sin i}{sin r} $

$\displaystyle \mu = \frac{sin45}{sin (90-C)} $

$\displaystyle \mu = \frac{sin45}{cosC} $

$\displaystyle \mu = \frac{1}{\sqrt{2} cosC} $

As , μ = 1/sinC

$\displaystyle \frac{1}{sinC} = \frac{1}{\sqrt{2} cosC} $

$\displaystyle tanC = \sqrt{2}$

$\displaystyle sinC = \frac{\sqrt{2}}{\sqrt{3}}$

$\displaystyle \frac{1}{\mu} = \frac{\sqrt{2}}{\sqrt{3}}$

$\displaystyle \mu = \sqrt{\frac{3}{2}}$