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# A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will:a) Become zero b) Become infinite c) Become small, but non-zero d) Remain unchanged

Last updated date: 17th Jul 2024
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Hint As per the question, we need to find the focal length of the convex lens which is dipped in a liquid with the same refractive index, we use the lens maker formula and calculate focal length and use the refractive index with respect to the refractive index of the liquid.

Complete Step by step solution
As we need to find out focal length of a convex lens, we will use lens maker formula, which is as below:
$\dfrac{1}{f} = \left( {\mu - 1} \right)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})......(1)$
Here, $f$ is focal length of convex lens, $\mu$is refractive index of convex lens and ${R_1}$, ${R_2}$ are radius of curvature of both surfaces of lens.
As per the question, the lens is dipped in a liquid whose refractive index. So, the lens maker formula will change because in this case the refractive index of the lens will be calculated with respect to the refractive index of the liquid.
Now, $\dfrac{1}{f} = \left( {\dfrac{{{\mu _1}}}{{{\mu _2}}} - 1} \right)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})......(2)$
Here, ${\mu _1}$ is refractive index of lens and ${\mu _2}$is refractive index of liquid
As given the refractive index of a lens is equal to the refractive index of liquid.
${\mu _1} = {\mu _2}$
So, the ratio of both refractive index will be$1$, we put it in equation $(2)$
$\dfrac{1}{f} = \left( {1 - 1} \right)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})$
$\dfrac{1}{f} = 0$
$\Rightarrow f = \infty$

Hence the focal length of the lens will be infinite, which is option b.

Note We should keep in mind that as per the question, we will use the lens maker formula to find out the focal length of the convex lens, but this lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. So the formula will change.