The critical angle for a denser-rarer interfa | vedclass

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(B) The relationship between the critical angle $i_C$,the refractive index $\mu$,and the speed of light in the two media is given by $\sin i_C = \frac

Vedclass · Accepted Answer

$2.12 	imes 10^8 \, m/s$

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(B) The relationship between the critical angle $i_C$,the refractive index $\mu$,and the speed of light in the two media is given by $\sin i_C = \frac{1}{\mu} = \frac{v_d}{v_r}$,where $v_d$ is the speed of light in the denser medium and $v_r$ is the speed of light in the rarer medium.Given $i_C = 45^{\circ}$ and $v_r = 3 	imes 10^8 \, m/s$.Substituting the values: $\sin 45^{\circ} = \frac{v_d}{3 	imes 10^8}$.Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\frac{1}{\sqrt{2}} = \frac{v_d}{3 	imes 10^8}$.Therefore,$v_d = \frac{3 	imes 10^8}{\sqrt{2}} \approx 2.12 	imes 10^8 \, m/s$.

The critical angle for a denser-rarer interface is $45^{\circ}$. The speed of light in the rarer medium is $3 \times 10^8 \, m/s$. The speed of light in the denser medium is:

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