Light travels in two media M_{1} and M_{2} wi | vedclass

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(A) The speed of light in a medium is given by $V = \frac{c}{n}$,where $c$ is the speed of light in vacuum and $n$ is the refractive index.Since $n =

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$	an^{-1}\left(\frac{3}{\sqrt{7}}ight)$

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(A) The speed of light in a medium is given by $V = \frac{c}{n}$,where $c$ is the speed of light in vacuum and $n$ is the refractive index.Since $n = \frac{c}{V}$,the refractive index is inversely proportional to the speed of light.For total internal reflection to occur,light must travel from a denser medium $(M_{1})$ to a rarer medium $(M_{2})$.The condition for the critical angle $i_{c}$ is $n_{1} \sin i_{c} = n_{2} \sin 90^{\circ}$.Thus,$\sin i_{c} = \frac{n_{2}}{n_{1}} = \frac{V_{1}}{V_{2}}$.Given $V_{1} = 1.5 	imes 10^{8} 	ext{ m/s}$ and $V_{2} = 2.0 	imes 10^{8} 	ext{ m/s}$.$\sin i_{c} = \frac{1.5 	imes 10^{8}}{2.0 	imes 10^{8}} = \frac{1.5}{2.0} = \frac{3}{4}$.If $\sin i_{c} = \frac{3}{4}$,then the opposite side is $3$ and the hypotenuse is $4$.The adjacent side is $\sqrt{4^{2} - 3^{2}} = \sqrt{16 - 9} = \sqrt{7}$.Therefore,$	an i_{c} = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{3}{\sqrt{7}}$.Hence,$i_{c} = 	an^{-1}\left(\frac{3}{\sqrt{7}}ight)$.

Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} \text{ m/s}$ and $2.0 \times 10^{8} \text{ m/s}$ respectively. The critical angle between them is.

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