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(A) According to Snell's Law,$\mu \sin 	heta_1 = \mu_1 \sin r_1$,where $r_1$ is the angle of refraction in the first block.If $	heta_1$ increases,$\

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increases

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(A) According to Snell's Law,$\mu \sin 	heta_1 = \mu_1 \sin r_1$,where $r_1$ is the angle of refraction in the first block.If $	heta_1$ increases,$\sin 	heta_1$ increases,so $\sin r_1$ must increase,meaning $r_1$ increases.At the interface between the two blocks,the angle of incidence is $r_1$ and the angle of refraction is $r_2$. By Snell's Law: $\mu_1 \sin r_1 = \mu_2 \sin r_2$.Since $\mu_1 \sin r_1$ has increased,$\mu_2 \sin r_2$ must also increase. Given $\mu_2$ is constant,$\sin r_2$ increases,so $r_2$ increases.At the final interface,the angle of incidence is $r_2$ and the angle of refraction is $	heta_2$. By Snell's Law: $\mu_2 \sin r_2 = \mu \sin 	heta_2$.Since $\mu_2 \sin r_2$ has increased,$\mu \sin 	heta_2$ must increase. Since $\mu$ is constant,$\sin 	heta_2$ increases,which means $	heta_2$ increases.

In the figure given below,a ray of light travelling in a medium of refractive index $\mu$ passes through two different connected rectangular blocks of refractive indices $\mu_1$ and $\mu_2$ $(\mu_2 > \mu_1)$. The angle of incidence $\theta_1$ is increased slightly. Then,the angle $\theta_2$ is

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