Question

For two structures namely $S_1$ with $n_1=\sqrt{45/4}$ and $n_2=3/2$, and $S_2$ with $n_1=8/5$ and $n_2=7/5$ and taking the refractive index of water to be 4/3 and that of air to be 1, the correct option is are

• A. NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$
• B. NA of $S_1$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is same as that of $S_2$ immersed in water
• C. NA of $S_1$ placed in air is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{4}{\sqrt{15}}$
• D. NA of $S_1$ placed in the air is the same as that of $S_2$ placed in water

Answer 1

Answer: (A), (C)

The correct options are
A NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$
C NA of $S_1$ placed in air is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{4}{\sqrt{15}}$

Using Snell's law$n_s\sin i=n_1\sin (90-i_c)$ where $n_s$ is the refractive index of the surrounding, $i_c$ is the critical angle for the medium for light from $n_1$ to $n_2$.

$NA=\sin i=\frac{n_1}{n_s}\sqrt{1-(\frac{n_2}{n_1}{)}^2}=\frac{1}{n_s}\sqrt{n_1^2-n_2^2}$

Option(A):

$N{A}_{S_1}=\frac{1}{\frac{4}{3}}\sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{9}{16},$ $N{A}_{S_2}=\frac{1}{\frac{16}{3\sqrt{15}}}\sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{9}{16}$

Option(B):

$N{A}_{S_1}=\frac{1}{\frac{\sqrt{15}}{6}}\sqrt{\frac{45}{16}-\frac{9}{4}}=\frac{9}{16},$ $N{A}_{S_2}=\frac{1}{\frac{\sqrt{15}}{6}}\sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{\sqrt{15}}{5}$

Option(C):

$N{A}_{S_1}=\frac{3}{4},$ $N{A}_{S_2}=\frac{3}{4}$

Option(D):

$N{A}_{S_1}=\frac{3}{4},$ $N{A}_{S_2}=\frac{3\sqrt{15}}{20}$