A convex lens is placed in water. Its focal length:
A) Increases
B) Decreases
C) Remains the same
D) None of the above
Answer
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Hint: The focal length of a lens is very much affected when immersed in water. The reason is that the angle of refraction is dependent on the relative difference between the indices of refraction of the incoming medium and the medium of the lens. The formula for the focal length is \[\dfrac{1}{f} = (\mu - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] , where $f$ is the focal length of the lens, $\mu $ is the ratio of the refractive indices of the material of the lens and the medium in which the lens is kept, and ${R_1}$ , ${R_2}$ are the radii of the curves of the lens.
Complete step by step solution:
Express the formula for the focal length of the lens
$\therefore \dfrac{1}{f} = (\mu - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ is a constant quantity for a lens
$\therefore \dfrac{1}{f} \propto (\mu - 1)$ or
$ \Rightarrow f \propto \dfrac{1}{{(\mu - 1)}}$
Now the change in focal length of the lens depends on the change in the $\mu $ . Let us see the change $\mu $ when the lens is immersed in water.
Suppose the refractive index of the lens is ${n_l}$ . We know that the refractive index of the air is about 1. Therefore when the lens is kept in the air then
$\therefore f \propto \dfrac{1}{{\dfrac{{{n_l}}}{1} - 1}}$
$ \Rightarrow f \propto \dfrac{1}{{{n_l} - 1}}$
When the lens is immersed in the water. And suppose the refractive index of the water is ${n_w}$
$\therefore f \propto \dfrac{1}{{\dfrac{{{n_l}}}{{{n_w}}} - 1}}$
We know that the value of ${n_w}$ is more than 1.
$\Rightarrow \mu = \dfrac{{{n_l}}}{{{n_w}}} < {n_l}$ , because ${n_w} > 1$
Thus the overall value of the denominator on the right-hand side will decrease. Now since the focal length is inversely proportional to the $(\mu - 1)$, therefore the focal length will increase.
Hence the focal length of the lens increases when it is immersed in the water.
Hence, Option A is correct.
Note: We compare the refraction from one medium to another by their refractive index. In the above question if we assume that the lens is immersed in a liquid. And suppose the refractive index of the liquid is equal to the refractive index of the lens. Then the lens will not change the path of the light and it will go through the lens in the same direction and path. If the refractive indices are the same then $\mu = 1$ . From the formula of the focal length, the focal length will be infinite. Which means the rays coming from infinite will converge at infinite. In other words, the converging power of the lens will be zero as $P = \dfrac{1}{f}$ .
Complete step by step solution:
Express the formula for the focal length of the lens
$\therefore \dfrac{1}{f} = (\mu - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ is a constant quantity for a lens
$\therefore \dfrac{1}{f} \propto (\mu - 1)$ or
$ \Rightarrow f \propto \dfrac{1}{{(\mu - 1)}}$
Now the change in focal length of the lens depends on the change in the $\mu $ . Let us see the change $\mu $ when the lens is immersed in water.
Suppose the refractive index of the lens is ${n_l}$ . We know that the refractive index of the air is about 1. Therefore when the lens is kept in the air then
$\therefore f \propto \dfrac{1}{{\dfrac{{{n_l}}}{1} - 1}}$
$ \Rightarrow f \propto \dfrac{1}{{{n_l} - 1}}$
When the lens is immersed in the water. And suppose the refractive index of the water is ${n_w}$
$\therefore f \propto \dfrac{1}{{\dfrac{{{n_l}}}{{{n_w}}} - 1}}$
We know that the value of ${n_w}$ is more than 1.
$\Rightarrow \mu = \dfrac{{{n_l}}}{{{n_w}}} < {n_l}$ , because ${n_w} > 1$
Thus the overall value of the denominator on the right-hand side will decrease. Now since the focal length is inversely proportional to the $(\mu - 1)$, therefore the focal length will increase.
Hence the focal length of the lens increases when it is immersed in the water.
Hence, Option A is correct.
Note: We compare the refraction from one medium to another by their refractive index. In the above question if we assume that the lens is immersed in a liquid. And suppose the refractive index of the liquid is equal to the refractive index of the lens. Then the lens will not change the path of the light and it will go through the lens in the same direction and path. If the refractive indices are the same then $\mu = 1$ . From the formula of the focal length, the focal length will be infinite. Which means the rays coming from infinite will converge at infinite. In other words, the converging power of the lens will be zero as $P = \dfrac{1}{f}$ .
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