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# A fish is near the center of a spherical water-filled fishbowl. A child stands in the air at a distance $2R$ ($R$ is the radius of curvature of the sphere) from the center of the bowl. At what distance from the center would the child's nose appear to the fish situated at the center:$(A)\;5R$ $(B)\;2R$$(C)\;3R$$(D)\;4R$

Last updated date: 15th Sep 2024
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Hint: By using the formula of refraction at the spherical refracting surface from rarer to the denser medium we can find the distance from the center where the child's nose appears.

Refraction at the spherical refracting surface from rarer to denser medium can be given by,
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
$\Rightarrow \dfrac{{\dfrac{4}{3}}}{v} - \dfrac{1}{{ - R}} = \dfrac{{\dfrac{4}{3} - 1}}{R}$
$\Rightarrow \dfrac{4}{{3v}} + \dfrac{1}{R} = \dfrac{1}{{3R}}$
$\Rightarrow \dfrac{4}{{3v}} = - \dfrac{2}{{3R}}$
$\Rightarrow v = - \dfrac{{4R}}{2} = - 2R$
The value is, $v = 2R$

Hence the right answer is option $(B) \Rightarrow 2R$.

$\dfrac{{\sin i}}{{\sin r}} = \mu$ [refractive index of the second medium with respect to the first medium]
Also $\mu = \dfrac{c}{v}$
$\mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}}$