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 - Physics MCQ AIEEE 2026 | ExamTest.live

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A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is and the fish is below the surface, the radius of this circle in cm is:

Solution & Step-by-step Explanation

The circular horizon is formed due to total internal reflection. The radius r of the circle at the surface is given by: r = h tan θ_{c} Where θ_{c} is the critical angle and sin θ_{c} = 1/ μ . tan θ_{c} = \frac{sin θ _{c}}{cos θ _{c}} = \frac{1/ μ}{1 - 1/ μ ^{2}} = \frac{1}{μ ^{2} - 1} r = \frac{h}{μ ^{2} - 1} Given h = 12 cm and μ = 4/3 : r = \frac{12}{( 4/3 ) ^{2} - 1} = \frac{12}{16/9 - 1} = \frac{12}{7/9} r = \frac{12}{7 /3} = \frac{36}{7} cm

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A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is and the fish is below the surface, the radius of this circle in cm is:

367

36/ 7

365

45

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