Question

There are two thin symmetrical lenses, one is converging with a refractive index ${n_1} = 1.70$ and the other is diverging with a refractive index ${n_2} = 1.50$. Both lenses have the same curvature radius of their surfaces equal to $R = 10cm$. The lenses are put close together and submerged into water. The focal length of the lens system is found to be $\dfrac{{100}}{x}cm$ in water. What is the value of $x$? (R.I of water $= \dfrac{4}{3}$).

Answer 1

Here, you are given two lenses out of which one is converging and the other one is diverging. The lenses are put close together and immersed in water and you are asked to find the focal length of the lens system. What you do is find the focal length of each lens individually in water and then use the formula to get the effective focal lens when two lenses are placed close to each other. In order to find the focal length of each lens in water given the refractive index, the radius of curvature and the refractive index of water, use the lens maker formula.

Complete step by step answer:
Let us consider a lens with radii of curvatures ${R_1}\& {R_2}$. For refraction at spherical surface, the general equation is given as $\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$, where ${\mu _1}\& {\mu _2}$ are refractive index of the mediums. For first refraction, that is at first face of lens, the equation will be $\dfrac{{{\mu _2}}}{{{v_1}}} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{{{R_1}}}$ and for the second refraction, that is the second face of lens, the equation will be $\dfrac{{{\mu _1}}}{v} - \dfrac{{{\mu _2}}}{{{v_1}}} = \dfrac{{{\mu _1} - {\mu _2}}}{{{R_2}}}$, here, ${v_1}$ is the distance of image formed due to first refraction, $v\& u$ are the final position and initial positions of the object. In general, the focal length is given by $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$.

Let us add the above two equations of refraction, we get,

\dfrac{{{\mu _2}}}{{{v_1}}} - \dfrac{{{\mu _1}}}{u} + \dfrac{{{\mu _1}}}{v} - \dfrac{{{\mu _2}}}{{{v_1}}} = \dfrac{{{\mu _2} - {\mu _1}}}{{{R_1}}} + \dfrac{{{\mu _1} - {\mu _2}}}{{{R_2}}} \\\ \Rightarrow{\mu _1}\left( {\dfrac{1}{v} - \dfrac{1}{u}} \right) = \left( {{\mu _2} - {\mu _1}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\\ \Rightarrow\dfrac{1}{v} - \dfrac{1}{u} = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\\ \Rightarrow\dfrac{1}{f} = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\\ $$ This is the lens maker formula. For first lens,

\dfrac{1}{{{f_1}}} = \left( {\dfrac{{1.7}}{{\dfrac{4}{3}}} - 1} \right)\left( {\dfrac{1}{{10}} - \dfrac{1}{{ - 10}}} \right) \\
\Rightarrow{f_1} = \dfrac{{200}}{{11}}cm \\ $$
For, second lens,

\dfrac{1}{{{f_2}}} = \left( {\dfrac{{1.5}}{{\dfrac{4}{3}}} - 1} \right)\left( {\dfrac{1}{{ - 10}} - \dfrac{1}{{10}}} \right) \\\ \Rightarrow{f_2} = - 40cm \\\ $$ When two lenses are placed together, the equivalent focal length of the combined lens system is given as $$\dfrac{1}{f} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$$. Therefore,

\dfrac{1}{f} = \dfrac{1}{{\dfrac{{200}}{{11}}}} + \dfrac{1}{{ - 40}} \\
\Rightarrow\dfrac{1}{f} = \dfrac{{11}}{{200}} - \dfrac{1}{{40}} \\
\Rightarrow\dfrac{1}{f} = \dfrac{{11 - 5}}{{200}} = \dfrac{6}{{200}} = \dfrac{3}{{100}} \\
\therefore f = \dfrac{{100}}{3}cm \\ $$
As it is given that the focal length found was $\dfrac{{100}}{x} cm$, by comparing with the focal length which we got, the value of $x$ will be $3$.

Therefore, the value of $x = 3$.

Note: Always keep the track of plus and minus signs wherever you tackle any refraction. In order to determine the sign of radius of curvature, you need to simply follow this rule, if you go in the direction of the light and the centre of curvature is also in that direction, take the sign of radius of curvature to be positive else, you have to take negative sign. Try to understand once more the signs used for the radius of curvature used in lens maker formula. Also keep in mind the lens maker formula and the equivalent focal length formula.