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# Two thin symmetrical lenses of different nature and of different material have equal radii of curvature $R = 15cm$ . The lenses are put close together and immersed in water $\left( {{\mu _w} = \dfrac{4}{3}} \right)$ . The focal length of the system in water is $30cm$. The difference between the refractive indices of the two lenses is A. $\dfrac{1}{2}$B. $\dfrac{1}{4}$C. $\dfrac{1}{3}$D. $\dfrac{3}{4}$

Last updated date: 06th Sep 2024
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Hint: Here, the lens maker formula needs to be applied in order to get the required solution. The lens maker's formula describes the relationship between a lens's focal length, material's refractive index, and the radii of curvature on both of its two surfaces. Manipulate the formula in terms of two lenses that consist of different materials by substituting the value of $\eta$. The refractive index of water is already given. The ratio of an electromagnetic wave's speed in a vacuum to its speed in another medium is known as the refractive index. The term "refractive index" describes how much a light ray bends as it travels through different media.

Formula used:
$\dfrac{{{\mu _w} - {\mu _1}}}{R} + \dfrac{{{\mu _2} - {\mu _w}}}{R} = \dfrac{1}{f}$

Complete step by step solution:
Before starting the solution, we write all the important things which are given in the question,
Radii of curvature of both symmetrical lenses, $R = 15cm$
Immersed in water, $\left( {{\mu _w} = \dfrac{4}{3}} \right)$
Focal length of water system = $30cm$
The lens maker formula is:
$\dfrac{1}{f}=\lgroup\eta-1\rgroup\lgroup\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\rgroup$
Now, we have to find the difference between the refractive indices of the two lenses,
As for which we have to use the following formula for further solutions,
$\dfrac{{{\mu _w} - {\mu _1}}}{R} + \dfrac{{{\mu _2} - {\mu _w}}}{R} = \dfrac{1}{f}$
By putting all the values in the above equation, we get the product as,
$\dfrac{\dfrac{4}{3}-\mu_{1}}{R}+\dfrac{\mu_{2}-\dfrac{4}{3}}{R}=\dfrac{1}{f}$
Now,
$\dfrac{\dfrac{4}{3}-\mu_{1}+\mu_{2}-\dfrac{4}{3}}{15}=\dfrac{1}{30}$
By doing a further calculation of the above equation, we get the equation as,
$\dfrac{\mu_{2}-\mu_{1}}{15}=\dfrac{1}{30}$
$\dfrac{{\Delta \mu }}{{15}} = \dfrac{1}{{30}}$
By rearranging the value we get the value of $\Delta \mu$,
$\Delta \mu = \dfrac{1}{2}$
Therefore, the correct answer for the difference between refractive indices of the two lenses is $\Delta \mu = \dfrac{1}{2}$.
Hence, the correct answer is A.

Note: It must be noted that The ratio of an electromagnetic wave's speed in a vacuum to its speed in another medium is known as the refractive index. The term "refractive index" describes how much a light ray bends as it travels through different media.