Here's a question in a paper based on refraction of light. I can't seem to solve it for some reason.
A man looks down at a fish length of 20 cm. His eye is 2m above the
surface of the water ($\mu = 4/3$) an the fish is 2m below the surface
shown in the figure. The ratio of angular width $\Delta\theta_1$ of
the fish as seen by the man in presence of water to the
$\Delta\theta_2$ in the absence of water is:
(A) 6/5 $\hspace{3cm}$ (B) 5/6 $\hspace{3cm}$ (C)7/8 $\hspace{3cm}$ (D)8/7
Here's what I've done:-
When there is no water present:
$$\frac{\Delta\theta_{air}}{2} = \sin^{-1}(10/400)$$
$$\therefore \Delta\theta_{air} = 2\sin^{-1}(1/40)$$
…from the triangle obtained by the eye, and half the fish.
When the water is present:
($P.S:$ the equation written below the diagram is wrong, I've just realised. I've written the correct version below)
Let the light rays at angle $\phi$ with the normal refract at the water surface and converge at the eye. Here the angle of refraction is $\theta$. So we have:
$$\sin\phi . \mu_{water} = \sin\theta . \mu_{air}$$
$$\therefore\frac{\sin\theta}{\sin\phi} = \frac{\mu_{water}}{\mu_{air}}$$
$$\therefore\frac{\sin\theta}{\sin\phi}=\frac{4}{3}$$
After which I am stuck and have no clue how to go about this. Can anyone help me?
Addendum: Will taking $\sin x = x$ help? $x$ is very small anyway, and $\sin x \approx x$ is turning out to be true ($x = 0.025$, $\sin(0.025) = 0.024997…$).
Addendum2: Another thing I figured out is that the water will increase the observed angular width, so the ratio is definitely $>1$. So we can safely rule out (B) and (C) which are $<1$.
Best Answer
The OP has calculated the $\theta _{air}$ correctly. To calculate the angle in the presence of water, we have:
$$2\tan{\theta}+ 2\tan{\phi}=0.1$$
Using small angle approximation:
$$\theta+\phi \approx \frac{1}{20}$$
However, from the Snell-Decartes law(as demonstrated by the OP) we have:
$$\frac{\sin{\theta}}{\sin\phi} \approx\frac{\theta}{\phi}\approx \frac{4}{3} \\ \Rightarrow \phi \approx \frac{3}{4}\theta \\ \Rightarrow \frac{7}{4}\theta=\frac{1}{20} \Rightarrow \theta=\frac{1}{35}$$
On the other hand, we have $\theta_{\text{air}}=\frac{1}{40}$; so the answeer would be:
$$\frac{\frac{1}{35}}{\frac{1}{40}}=\frac{8}{7}$$