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Step-by-Step Solution
Step 1: Express the focal length of the plano-convex lens
For a plano-convex lens (one surface plane, the other spherical) made of material with refractive index
$ \mu_1 $, the lens maker’s formula simplifies to:
$ \frac{1}{f_1} = \bigl(\mu_1 - 1 \bigr) \left(\frac{1}{R} - \frac{1}{\infty}\right). $
Here, $ R $ is the radius of curvature of the spherical face and the plane side has an infinite radius of curvature. Thus,
$ \frac{1}{f_1} = (\mu_1 - 1)\frac{1}{R}. $
Step 2: Express the focal length of the plano-concave lens
For a plano-concave lens (one surface plane, the other spherical) made of material with refractive index
$ \mu_2 $, the lens maker’s formula becomes:
$ \frac{1}{f_2} = \bigl(1 - \mu_2 \bigr)\left(\frac{1}{R} - \frac{1}{\infty}\right) $
or equivalently
$ \frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-\!R}\right). $
Since for a concave surface the radius is taken as $ -R $, we can write:
$ \frac{1}{f_2} = -(\mu_2 - 1)\frac{1}{R}
\quad\Longrightarrow\quad
\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{-1}{R}\right). $
For simplicity, we write it as:
$ \frac{1}{f_2} = \frac{\mu_2 - 1}{-\,R}. $
Step 3: Relate the focal lengths given the condition $ f_1 = 2\,f_2 $
We are given that $ f_1 = 2f_2 $, which implies $ \frac{1}{f_1} = \frac{1}{2f_2} $. Using the expressions for
$ \frac{1}{f_1} $ and $ \frac{1}{f_2} $ derived above:
$ \frac{1}{f_1} = \frac{\mu_1 - 1}{R},
\quad
\frac{1}{f_2} = \frac{\mu_2 - 1}{-\,R}. $
Substituting $ \frac{1}{f_1} = \frac{1}{2f_2} $ gives:
$ \frac{\mu_1 - 1}{R} = \frac{1}{2} \left(\frac{\mu_2 - 1}{-\,R}\right). $
Step 4: Simplify the relation to find $ 2\mu_1 - \mu_2 = 1 $
Simplifying the above:
$ \frac{\mu_1 - 1}{R} = \frac{1}{2} \cdot \frac{\mu_2 - 1}{-R}
\quad\Longrightarrow\quad
\mu_1 - 1 = \frac{1}{2} (-(\mu_2 - 1))
\quad\Longrightarrow\quad
\mu_1 - 1 = -\frac{1}{2}(\mu_2 - 1). $
Multiply both sides by $ -2 $ to remove the fraction and simplify:
$ -2(\mu_1 - 1) = (\mu_2 - 1)
\quad\Longrightarrow\quad
-2\mu_1 + 2 = \mu_2 - 1. $
Rearrange terms to get:
$ 2\mu_1 - \mu_2 = 1. $
Final Relation
Thus, the relationship between $ \mu_1 $ and $ \mu_2 $ is:
$ \,2\mu_1 - \mu_2 = 1. $
