Question

A fish, looking up through the water, sees the outside world contained in a circular horizon. If the refractive index of water is \(4/3\) and the fish is \(12\,cm\) below the surface of water, the radius of the circle in centimetre is

• A. \(\dfrac{12\times 3}{\sqrt{5}}\)
• B. \(12\times 3\times \sqrt{5}\)
• C. \(\dfrac{12\times 3}{\sqrt{7}}\)
• D. \(12\times 3\times \sqrt{7}\)

Hint:

For fish in water, horizon radius is \(r=h\tan C\), where \(\sin C=\dfrac{1}{\mu}\). Always compute \(\tan C\) from sine.

Answer 1

The Correct Option is C

Step 1: Concept of circular horizon (Snell’s law).
A fish can see outside world only for rays that emerge from water surface.
The limiting ray corresponds to the critical angle \(C\).
Step 2: Find critical angle for water-air interface.
\[ \sin C = \frac{1}{\mu} \] Here \(\mu = \frac{4}{3}\).
\[ \sin C = \frac{1}{4/3} = \frac{3}{4} \] Step 3: Use geometry to find radius.
Fish is at depth \(h=12\,cm\).
At surface, circular radius \(r\) is:
\[ r = h\tan C \] Step 4: Find \(\tan C\).
\[ \sin C = \frac{3}{4} \Rightarrow \cos C = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \] \[ \tan C = \frac{\sin C}{\cos C} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}} \] Step 5: Calculate radius.
\[ r = 12 \times \frac{3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \] Final Answer: \[ \boxed{\dfrac{12\times 3}{\sqrt{7}}} \]