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# A convex lens of focal length 0.2 m & made of glass $\left( {\mu = 1.5} \right)$  is immersed in water $\left( {\mu = 1.33} \right)$ . Find the change in the focal length of the lens.

Last updated date: 21st Jun 2024
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Hint: We have to use the lens maker formula $\left( {\dfrac{1}{f} = \left( {\dfrac{\mu}{\mu _m} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)} \right)$  here to find the focal lengths in air and water respectively and then we can compare them to find the change in focal length.

To  solve this question we need the ratio between the focal length of lens in air and in water using which we can find the focal length in water itself.
Lens maker’s formula is the relation between the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces. It is used by lens manufacturers to make the lenses of particular power from the glass of a given refractive index.
We know, From the Lens Maker formula-
$\left( {\dfrac{1}{f} = \left( {\dfrac{\mu}{\mu _m} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)} \right)$
where,
$f =$ focal length of lens
$\mu =$ refractive index of  lens  (1.5 for glass)
${\mu _m} =$ refractive index of the medium in which the lens is placed.
${R_1},{R_2}$ are the radii of curvature of two faces of the lens.
We know that the refractive index of air is 1.
Hence, we find the focal length of lens in air (say ${f_a}$ ) as,
$\dfrac{1}{{{f_a}}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Also, the refractive index of water is 1.33. So, the focal length of lens in water (say ${f_w}$ ) is
$\dfrac{1}{f} = \left( \dfrac{{1.5 - 1.33}}{1.33} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
On dividing the above two equations, we get
$\dfrac{{{f_w}}}{{{f_a}}} = \dfrac{{0.128}}{{1.5 - 1.33}}$
$\Rightarrow \dfrac{{{f_w}}}{{{f_a}}} = \dfrac{{0.128}}{{0.17}} = 0.75$
$\Rightarrow {f_w} = 0.75 \times {f_a}$
According to the question, the focal length of lens in air, ${f_a} = 0.2m$
Substituting the above value in, we get the focal length of water,
$\therefore{f_w} = 0.2 \times 0.75 = 0.15m$

Hence, the correct answer is that the change in focal length is 0.05m.

Note: We have to keep in mind that this question can be solved even without knowing the radii of curvature but that’s not the case in all questions, so it has to be taken in consideration. Also, the radius of curvature for a flat face is taken to be infinity, $R \to \infty \Rightarrow \dfrac{1}{R} \to 0$ , so the term gets cancelled away.