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(D) The speed of light in medium $B$ $(v_B)$ is $1.5 	imes 10^{10} \, cm/s$ and in medium $A$ $(v_A)$ is $2.0 	imes 10^{10} \, cm/s$.Since $v_B < v_

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$	heta > \sin^{-1}\left(\frac{3}{4}ight)$

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(D) The speed of light in medium $B$ $(v_B)$ is $1.5 	imes 10^{10} \, cm/s$ and in medium $A$ $(v_A)$ is $2.0 	imes 10^{10} \, cm/s$.Since $v_B The critical angle $i_c$ for light traveling from a denser medium to a rarer medium is given by $\sin i_c = \frac{n_r}{n_d} = \frac{v_d}{v_r}$.Here,$v_d = v_B = 1.5 	imes 10^{10} \, cm/s$ and $v_r = v_A = 2.0 	imes 10^{10} \, cm/s$.$\sin i_c = \frac{1.5 	imes 10^{10}}{2.0 	imes 10^{10}} = \frac{1.5}{2.0} = \frac{3}{4}$.Therefore,$i_c = \sin^{-1}\left(\frac{3}{4}ight)$.For total internal reflection $(TIR)$ to occur,the angle of incidence $	heta$ must be greater than the critical angle $i_c$.Thus,$	heta > \sin^{-1}\left(\frac{3}{4}ight)$.

The speed of light in media $A$ and $B$ are $2.0 \times 10^{10} \, cm/s$ and $1.5 \times 10^{10} \, cm/s$ respectively. $A$ ray of light enters from medium $B$ to $A$ at an incident angle $\theta$. If the ray suffers total internal reflection,then:

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