Question

Two Plano-concave lenses (1 and 2) of glass of refractive index 1.5 have radii of curvature 25cm and 20cm. They are placed in contact with their curved surface towards each other and the space between them is filled with liquid of refractive index 4/3. Then the combination is

a) Convex lens of focal length 70cm.
b) Concave lens of focal length 70cm.
c) Concave lens of focal length 66.6cm
d) Convex lens of focal length 66.6cm

Answer 1

Hint:Here we need to consider the liquid inside the two lenses to be a third lens of refractive index 4/3. Find out the focal length of each lens separately including the liquid inside the lenses and add them together to get the final focal length.

Complete step by step Solution:
The formula for focal length is:
$\dfrac{1}{F} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ ;
Here:
F = Focal Length;
n = Refractive Index;
R = Radius of curvature;

Step 1:
The equivalent focal length would be the combination of three focal lengths.
$\dfrac{1}{F} = \dfrac{1}{{{F_1}}} + \dfrac{1}{{{F_2}}} + \dfrac{1}{{{F_3}}}$;
The combination for both the process:
$\dfrac{1}{{{F_1}}} = \left( {{n_1} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$;
$ \Rightarrow \dfrac{1}{{{F_1}}} = \left( {\dfrac{3}{2} - 1} \right)\left( {\dfrac{1}{\infty } - \dfrac{1}{{25}}} \right)$;
Do the necessary calculations:
$ \Rightarrow \dfrac{1}{{{F_1}}} = \left( {\dfrac{1}{2}} \right) \times \left( { - \dfrac{1}{{25}}} \right)$;
$ \Rightarrow \dfrac{1}{{{F_1}}} = - \dfrac{1}{{50}}$;
The focal length of lens 1 is $\dfrac{1}{{{F_1}}} = \left( { - \dfrac{1}{{50}}} \right)$;
Similarly, for the second lens:
$\dfrac{1}{{{F_2}}} = \left( {\dfrac{3}{2} - 1} \right)\left( {\dfrac{1}{{ - 20}} - \dfrac{1}{\infty }} \right)$;
Do the needed calculation:
\[ \Rightarrow \dfrac{1}{{{F_2}}} = \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{ - 20}}} \right)\];
\[ \Rightarrow \dfrac{1}{{{F_2}}} = \left( { - \dfrac{1}{{40}}} \right)\];
Similarly, we can do this for third focal length:
$\dfrac{1}{{{F_3}}} = \left( {\dfrac{4}{3} - 1} \right)\left( {\dfrac{1}{{25}} + \dfrac{1}{{20}}} \right)$;
Do the necessary calculation:
$ \Rightarrow \dfrac{1}{{{F_3}}} = \left( {\dfrac{1}{3}} \right)\left( {\dfrac{9}{{100}}} \right)$;
$ \Rightarrow \dfrac{1}{{{F_3}}} = \dfrac{3}{{100}}$

Step 3:
For the net focal length add all the three focal lengths together.
$\dfrac{1}{F} = \dfrac{1}{{{F_1}}} + \dfrac{1}{{{F_2}}} + \dfrac{1}{{{F_3}}}$;
$ \Rightarrow \dfrac{1}{F} = - \dfrac{1}{{50}} - \dfrac{1}{{40}} + \dfrac{3}{{100}}$;
Solve, the above equation for F:
$ \Rightarrow \dfrac{1}{F} = - \dfrac{{15}}{{1000}}$;
$ \Rightarrow F = - \dfrac{{1000}}{{15}}$;
The net focal length comes out to be
$ \Rightarrow F = - 66.66cm$;

Final Answer:Option “c” is correct. Two Plano-concave lenses are placed in contact with their curved surface towards each other and the space between them is filled with liquid of refractive index 4/3. Then the combination is a Concave lens of focal length 66.6cm.

Note:Here the option is a concave lens because the focal length lies behind the lens, if the focal length would have lied in front of the lens then the lens would be a convex lens. Here, for each lens the other part for the radius of curvature goes to infinity.