2
$\begingroup$

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

fish


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 $<1$.

$\endgroup$
  • $\begingroup$ You will indeed need to use the small angles approximation. $\endgroup$ – Ali Jul 27 '13 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. $\endgroup$ – mikhailcazi Jul 27 '13 at 12:43
  • $\begingroup$ Okay, I'll post an answer. $\endgroup$ – Ali Jul 27 '13 at 12:46
4
$\begingroup$

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:

$$\frac{\frac{1}{35}}{\frac{1}{40}}=\frac{8}{7}$$

$\endgroup$
  • $\begingroup$ Could you please explain how you got $\tan(\theta /2) + \tan(\phi / 2) = 0.1/4$? $\endgroup$ – mikhailcazi Jul 27 '13 at 13:33
  • $\begingroup$ @mikhailcazi I'll add a figure for that. $\endgroup$ – Ali Jul 27 '13 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 '13 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.