Question

Two identical thin plano-convex lenses of refractive index n are silvered, one on the plane side and the other on the convex side. The ratio of their focal length is
A. $\frac{n}{n-1}$
B. $\frac{n-1}{n}$
C. $\frac{n+1}{n}$
D. $n$

Answer 1

Hint:We have two identical thin plano convex lenses. Both have refractive index n. It is given that the plane side of one plano convex lens is silver and the convex side of another plano convex lens is painted. According to the question we have to find the ratio of focal length of two plano convex lenses.


Formula used:

Here we have a combination of a mirror and lens. So, we use equation for power which is:
$P=2{{P}_{l}}-{{P}_{m}}$
Where P is the power of combination, Pl is the power of lens and Pm is the power of mirror.
To simplify this equation, we use a relation connecting focal length and power. That is,
$P=\frac{1}{f}$
Where f is the focal length and P is the power.



Complete answer:
We have the equation connecting focal lengths as:
$\frac{1}{f}=\frac{2}{{{f}_{l}}}+\frac{2}{{{f}_{m}}}$
Where ${{f}_{l}}$ is the focal length of the lens and ${{f}_{m}}$ is the focal length of the mirror.
We know that when the plano convex lens is painted on the convex side, then their focal length will be
${{f}_{m}}=\frac{R}{2}$
Then we can write:
$\frac{1}{{{f}_{l}}}=\frac{n-1}{R}$
So, we can write it as:
${{f}_{2}}=\frac{R}{2n}$
If the plane side of this plano convex lens is painted, then their focal length changes to:
${{f}_{m}}=0$
Then we have:
${{f}_{l}}=\frac{n-1}{R}$
In total focal length this time will be:
${{f}_{1}}=\frac{R}{2(n-1)}$
Then the required ratio will be
$\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{n}{n-1}$
Therefore, the answer is option (A)



Note:Remember that one side is painted silver means that side of the plano convex lens acts as a mirror. That is here it is a combination of lens and mirror. Remember that the power of a mirror is always negative and we should consider rays passing through the lens two times.