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Geometrical Optics question

2005 · Shift 0 · Q120
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Geometrical Optics question

2005 · Shift 0 · Q120

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 43{4 \over 3}34​ and the fish is 12cm12cm12cm below the surface, the radius of this circle in cmcmcm is
  1. A
    367{{36} \over {\sqrt 7 }}7​36​
  2. B
    367{36\sqrt 7 }367​
  3. C
    45{4\sqrt 5 }45​
  4. D
    365{36\sqrt 5 }365​
View written solutionFree

Correct answer: A

  1. Physical idea: circular horizon seen by the fish

A fish under water can see the outside world only through rays that emerge from air into water and reach its eye.

The limiting rays correspond to the critical angle at the water–air surface. These rays form a cone, and its intersection with the water surface is a circle. The fish sees the outside world within this circular boundary.

  1. Find the critical angle

For water to air,

sin⁡c=nairnwater=14/3=34.\sin c = \frac{n_{\text{air}}}{n_{\text{water}}} = \frac{1}{4/3} = \frac{3}{4}.sinc=nwater​nair​​=4/31​=43​.

So,

sin⁡c=34.\sin c = \frac{3}{4}.sinc=43​.

Then,

cos⁡c=1−sin⁡2c=1−916=716=74.\cos c = \sqrt{1-\sin^2 c} = \sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt7}{4}.cosc=1−sin2c​=1−169​​=167​​=47​​.

Hence,

tan⁡c=sin⁡ccos⁡c=3/47/4=37.\tan c = \frac{\sin c}{\cos c} = \frac{3/4}{\sqrt7/4} = \frac{3}{\sqrt7}.tanc=coscsinc​=7​/43/4​=7​3​.
  1. Relate radius of circular horizon to fish depth

If the fish is at depth h=12 cmh=12\,\text{cm}h=12cm, then from the right triangle,

r=htan⁡c.r = h\tan c.r=htanc.

Therefore,

r=12⋅37=367 cm.r = 12\cdot \frac{3}{\sqrt7} = \frac{36}{\sqrt7}\,\text{cm}.r=12⋅7​3​=7​36​cm.
  1. Match with options
r=367 cm\boxed{r=\frac{36}{\sqrt7}\,\text{cm}}r=7​36​cm​

This corresponds to Option A.

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