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: diag
($P.S:$ 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 $<1$.

  • $\begingroup$ You will indeed need to use the small angles approximation. $\endgroup$
    – Ali
    Jul 27, 2013 at 12:38
  • $\begingroup$ @Ali Okay, cool, so I get $\Delta\theta_{air} = 1/20$. But I'm stuck trying to find the angular width when the water is present. $\endgroup$ Jul 27, 2013 at 12:43
  • $\begingroup$ Okay, I'll post an answer. $\endgroup$
    – Ali
    Jul 27, 2013 at 12:46

1 Answer 1


The OP has calculated the $\theta _{air}$ correctly. To calculate the angle in the presence of water, we have:

$$2\tan{\theta}+ 2\tan{\phi}=0.1$$

enter image description here

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:


  • $\begingroup$ Could you please explain how you got $\tan(\theta /2) + \tan(\phi / 2) = 0.1/4$? $\endgroup$ Jul 27, 2013 at 13:33
  • $\begingroup$ @mikhailcazi I'll add a figure for that. $\endgroup$
    – Ali
    Jul 27, 2013 at 13:34
  • $\begingroup$ @mikhailcazi my $\theta$ was your $2 \theta$! I will correct the answer corresponding to your own notation. $\endgroup$
    – Ali
    Jul 27, 2013 at 13:51

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