Question

In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is R. The focal length of this system is.
A. \[f = \dfrac{{{\mu _3}R}}{{{\mu _3} - {\mu _1}}}\]
B. \[f = \dfrac{{{\mu _2}R}}{{{\mu _1} - {\mu _2}}}\]
C. \[f = \dfrac{{{\mu _3}}}{{{\mu _2} - {\mu _1}}}\]
D. \[f = \dfrac{{{\mu _1}R}}{{{\mu _1} - {\mu _2}}}\]

Answer 1

Hint: The focal length can be defined as the distance between the centre of the lens (as it is a curve in shape) and the focus of it. So here it is given two surfaces, firstly find the refractive index of the first surface and the second surface and then add both to get the focal length.

Complete step by step answer:
Given:
Let the radius of the curvature for the surface is R.
Let the incident of the light one a thin lens is\[f\].
Then the refractive index of the first surface is \[{\mu _1}\].
The refractive index of the second surface is \[{\mu _2}\]
The refractive index of the other surface is \[{\mu _3}\]
The distance between the image to the first surface of the lens is \[{\nu _1}\]
The distance between the image to the second surface of the lens is \[{\nu _2}\]
Then the refraction at the first surface involves
\[{\mu _2}\],\[{\mu _1}\]and \[{\nu _1}\]then it can be written as
\[\dfrac{{{\mu _2}}}{{{\nu _1}}} - \dfrac{{{\mu _1}}}{{ - \infty }} = \dfrac{{{\mu _2} - {\mu _1}}}{R}\]-- (1)
Now, taking the refraction at the second surface, it involves\[{\mu _2}\],\[{\mu _3}\],\[{\nu _1}\]and \[{\nu _2}\], then the equation will be
\[\dfrac{{{\mu _3}}}{{{\nu _2}}} - \dfrac{{{\mu _2}}}{{{\nu _1}}} = \dfrac{{{\mu _3} - {\mu _2}}}{R}\]--(2)
Now, adding the both equations 1 and 2, then
\[\dfrac{{{\mu _2}}}{{{\nu _1}}} - \dfrac{{{\mu _1}}}{{ - \infty }} + \dfrac{{{\mu _3}}}{{{\nu _2}}} - \dfrac{{{\mu _2}}}{{{\nu _1}}} = \dfrac{{{\mu _2} - {\mu _1}}}{R} + \dfrac{{{\mu _3} - {\mu _2}}}{R}\]
Now solving the above equation we will get
\[\dfrac{{{\mu _3}}}{{{\nu _2}}} - \dfrac{{{\mu _3} - {\mu _1}}}{R}\]
\[{\nu _2} = \dfrac{{R{\mu _3}}}{{{\mu _3} - {\mu _1}}}\]
Here, \[{\nu _2}\]will be the focal length then, replacing the \[{\nu _2}\]as\[f\], then the equation becomes
\[f = \dfrac{{R{\mu _3}}}{{{\mu _3} - {\mu _1}}}\]
Therefore, the focal length of the system is\[f = \dfrac{{R{\mu _3}}}{{{\mu _3} - {\mu _1}}}\]

So, the correct answer is “Option A”.

Note:
here we have to take the ratios of the refracting length and the focal length in the both surfaces. So remember this while taking those ratios. While adding the both equations, do not confuse between the focal length and the refractive index.