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A fish in near the centre of a spherical fish bowl filled with water of refractive index $\frac{4}{3} .$ A child stands in air at a distance 2R (R is the radius of curvature of the sphere) from the centre of the bowl. At what distance from the centre would the child’s nose appear to the fish situated at the centre

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a

R

b

2R

c

3R

d

4R

answer is C.

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Detailed Solution

Here nose of the boy is the object and fish is observer. Using refraction formula, we have
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$ where $u = + R, μ_{1} = 1, μ_{2} = \frac{4}{3}, R = - R$
$\Rightarrow \frac{4}{3 v} - \frac{1}{R} = \frac{\frac{4}{3} - 1}{- R} \Rightarrow v = + 2 R$
Thus, the image of child’s nose will appear at a distance 3R from the centre of the bowl. The ray diagram is shown in below figure.
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A fish in near the centre of a spherical fish bowl filled with water of refractive index 43. A child stands in air at a distance 2R (R is the radius of curvature of the sphere) from the centre of the bowl. At what distance from the centre would the child’s nose appear to the fish situated at the centre