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Q. Light travels in two media $M_1$ and $M_2$ with speeds $1.5 \times 10^8 \,ms ^{-1}$ and $2.0 \times 10^8 \,ms ^{-1}$ respectively. The critical angle between them is:

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Solution:

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$ v =\frac{ c }{ n } $
$ n _{ d } \sin i _{ c }= n _{ r } \sin 90^{\circ} $
$ \sin i _{ c }=\frac{ n _{ r }}{ n _{ d }}=\frac{ v _{ d }}{ v _{ r }}$
$ \sin i _{ c }=\frac{1.5 \times 10^8}{2 \times 10^8}=\frac{1.5}{2} $
$ \sin i _{ c }=\frac{3}{4}$
$ \tan i _{ c }=\frac{3}{\sqrt{4^2-3^2}} \Rightarrow \frac{3}{\sqrt{7}}$
$ i _{ c }=\tan ^{-1}\left(\frac{3}{\sqrt{7}}\right)$