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Byju's Answer

Standard XII

Physics

Combinations of Lenses & Mirrors

A fish is nea...

Question

A fish is near the center of a spherical water filled $(μ = 4 / 3)$ fish bowl. A child stands in air at a distance $2 R$ ( $R$ 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 :

5 R

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2 R

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3 R

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4 R

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Solution

The correct option is C $3 R$
We use the formula for different refractive indices,

$\frac{1}{v} - \frac{4}{3 u} = \frac{- \frac{4}{3} + 1}{R}$

Substituting, $u = 2 R$

$\frac{1}{v} - \frac{2}{3 R} = \frac{- 1}{3 R}$

This gives the value of $v$ as $3 R$

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A fish is near the center of a spherical water filled (μ=4/3) fish bowl. A child stands in 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 :

The correct option is C 3RWe use the formula for different refractive indices,1v−43u=−43+1RSubstituting, u=2R1v−23R=−13RThis gives the value of v as 3R

A fish is near the center of a spherical water filled $(μ = 4 / 3)$ fish bowl. A child stands in air at a distance $2 R$ ( $R$ 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 :

The correct option is C $3 R$
We use the formula for different refractive indices,

$\frac{1}{v} - \frac{4}{3 u} = \frac{- \frac{4}{3} + 1}{R}$

Substituting, $u = 2 R$

$\frac{1}{v} - \frac{2}{3 R} = \frac{- 1}{3 R}$

This gives the value of $v$ as $3 R$