Light travels in two media M1 and M2 with speeds 1.5×108ms−1 and 2×106ms−1 respectively. The critical angle between them is

ndsinic=nrsin90° (From Snell law) sinic3=nrnd=VdVr (V=cn) sinic=1.5×1082×108=34 Tanic=34n−32=37Tanic=37 ic=Tan−1(37)

Light travels in two media M1 and M2 with speeds 1.5×108ms−1 and 2×106ms−1 respectively. The critical angle between them is

ndsinic=nrsin90° (From Snell law) sinic3=nrnd=VdVr (V=cn) sinic=1.5×1082×108=34 Tanic=34n−32=37Tanic=37 ic=Tan−1(37)

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Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} m s^{- 1}$ and $2 \times 10^{6} m s^{- 1}$ respectively. The critical angle between them is

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$\tan^{- 1} (\frac{2}{3})$

$\cos^{- 1} (\frac{3}{4})$

$\sin^{- 1} (\frac{2}{3})$

$\tan^{- 1} (\frac{3}{\sqrt{7}})$

answer is A.

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Detailed Solution

$n_{d} \sin i_{c} = n_{r} \sin 90 °$ (From Snell law)
$\sin i_{c}^{3} = \frac{n_{r}}{n_{d}} = \frac{V_{d}}{V_{r}} (V = \frac{c}{n})$ $\sin i_{c} = \frac{1.5 \times 10^{8}}{2 \times 10^{8}} = \frac{3}{4}$ $T a n i_{c} = \frac{3}{\sqrt{4^{n} - 3^{2}}} = \frac{3}{\sqrt{7}}$