Question

A plano-convex lens acts like a concave mirror of \[28\,{\text{cm}}\] focal length when its plane surface is silvered and like a concave mirror of \[10\,{\text{cm}}\] focal length when curved surface is silvered. What is the refractive index of the material of the lens?
A. 1.50
B. 1.55
C. 1.60
D. 1.65

Answer 1

Hint: Use the equation for the focal length of the plano-convex lens. Also use the equation of lens maker’s formula for the plano-convex lens. These equations will give the relation between the focal lengths of the concave mirrors formed and the refractive index of the lens.

Formula used:
The equation for the focal length of the plano-convex lens is
\[\dfrac{1}{{{f_1}}} = \dfrac{2}{f} + \dfrac{1}{{{f_m}}}\] …… (1)
Here, \[{f_1}\] is the focal length of the plano-convex lens, \[f\] is the focal length of the convex surface of the lens and \[{f_m}\] is the focal length of the mirror surface.
The lens maker’s formula for the plano-convex lens is
\[\dfrac{1}{f} = \left( {\mu - 1} \right)\dfrac{1}{R}\] …… (2)
Here, \[f\] is the focal length of the convex surface of the plano-convex lens, \[\mu \] is the refractive index of the lens and \[R\] is its radius of curvature.

Complete step by step answer:
We can see that when the plane surface of a plano-convex lens is silvered, it becomes a concave mirror of focal length \[28\,{\text{cm}}\].
\[{f_1} = 28\,{\text{cm}}\]

The focal length of the plane surface is infinite.
\[{f_m} = \infty \]

Let us calculate the focal length \[f\] of the concave mirror formed by silvering of the plane surface of the plano-convex mirror.

Substitute \[28\,{\text{cm}}\] for \[{f_1}\] and \[\infty \] for \[{f_m}\] in equation (1).
\[\dfrac{1}{{28\,{\text{cm}}}} = \dfrac{2}{f} + \dfrac{1}{\infty }\]
\[ \Rightarrow \dfrac{1}{{28\,{\text{cm}}}} = \dfrac{2}{f} + 0\]
\[ \Rightarrow f = 56\,{\text{cm}}\]

Hence, the focal length of the concave mirror formed by silvering the plane surface is \[56\,{\text{cm}}\].
As the focal length of the plane surface is infinite, equation (1) becomes
\[f = 2{f_1}\]

Substitute \[2{f_1}\] for \[f\] in equation (2).
\[\dfrac{1}{{2{f_1}}} = \left( {\mu - 1} \right)\dfrac{1}{R}\]
\[ \Rightarrow {f_1} = \dfrac{R}{{2\left( {\mu - 1} \right)}} = 28\,{\text{cm}}\] …… (3)

The focal length \[{f_2}\] for the second mirror formed is \[10\,{\text{cm}}\].
\[ \Rightarrow {f_2} = \dfrac{R}{{2\mu }} = 10\,{\text{cm}}\] …… (4)

Divide equation (3) by equation (4).
\[ \Rightarrow \dfrac{{\dfrac{R}{{2\left( {\mu - 1} \right)}}}}{{\dfrac{R}{{2\mu }}}} = \dfrac{{28\,{\text{cm}}}}{{10\,{\text{cm}}}}\]
\[ \Rightarrow \dfrac{\mu }{{\mu - 1}} = 2.8\]
\[ \Rightarrow \mu = 2.8\mu - 2.8\]
\[ \Rightarrow \mu = 1.55\]

Therefore, the refractive index of the lens is \[1.55\].

So, the correct answer is “Option B”.

Note:
The focal length of the plane surface of the lens is infinite and hence, the ratio of one with infinity is zero. Hence, the radius of curvature for the plane surface of the mirror is also zero. Hence, the lens maker’s formula is modified as shown in equation (2).