Question

A plano-convex lens when silvered at its plane surface is equivalent to a concave mirror of focal length 28cm. when its curved surface is silvered and not the plane surface, it is equivalent to a concave mirror of focal length 10 cm, then the refractive index of the material of the lens is
(A) $\dfrac{9}{{14}}$
(B) $\dfrac{{14}}{9}$
(C) $\dfrac{{17}}{9}$
(D) None of these

Answer 1

Using the net focal length of the two circumstances, we need to find the radius of curvature of the silvered surface which is twice the focal length of the silvered surface. Then we can use the lens maker equation to calculate the refractive index of the medium.

Formula used: In this solution we will be using the following formulae;
$\dfrac{1}{{{f_{net}}}} = \dfrac{2}{{{f_l}}} + \dfrac{1}{{{f_s}}}$ where ${f_{net}}$ is the net focal length of the lens acting as a mirror, ${f_u}$ is the focal length of the lens (as a lens), ${f_s}$ is the focal length of the silvered surface.
$R = 2f$ where $R$ is the radius of curvature of a particular surface and $f$ is the focal length of the surface.
$\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $f$ in this case is the focal length of an entire block, $n$ is the refractive index of the block, ${R_1}$ is the radius of curvature of one of the surfaces of the block and ${R_2}$ is the radius for the other surface.

Complete Step-by-Step solution
To calculate the refractive index, we need the focal length of the curved surface.
When a lens is being silvered on one side to acts as a mirror, the net focal length of the entire system is given by
$\dfrac{1}{{{f_{net}}}} = \dfrac{2}{{{f_l}}} + \dfrac{1}{{{f_s}}}$ where ${f_{net}}$ is the net focal length of the lens acting as a mirror, ${f_u}$ is the focal length of the lens (as a lens), ${f_s}$ is the focal length of the silvered surface.
Hence for when it is a concave mirror,
$\dfrac{1}{{28}} = \dfrac{2}{{{f_l}}} + \dfrac{1}{\infty }$ . since the silvered surface is flat, it has a focal length of infinity.
Hence,
${f_u} = 2 \times 28 = 56cm$ .
Now, in the second case, the curved surface is silvered. Hence, we have
$\dfrac{1}{{10}} = \dfrac{2}{{{f_l}}} + \dfrac{1}{{{f_s}}}$
From above, ${f_l} = 56cm$
$\dfrac{1}{{10}} = \dfrac{2}{{56}} + \dfrac{1}{{{f_s}}}$
Calculating for ${f_s}$ , we have
${f_s} = \dfrac{{140}}{9}cm$
Then the curved surface is as above.
Now, $R = 2f$ where $R$ is the radius of curvature of a particular surface and $f$ is the focal length of the surface. Then,
${R_c} = 2\left( {\dfrac{{140}}{9}} \right) = \dfrac{{280}}{9}cm$
But the focal length of the entire block (as a lens) can be given as
$\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $f$ in this case is the focal length of an entire block, $n$ is the refractive index of the block, ${R_1}$ is the radius of curvature of one of the surfaces of the block and ${R_2}$ is the radius for the other surface.
Then, inserting all known values, we have
$\dfrac{1}{{56}} = \left( {n - 1} \right)\left( {\dfrac{9}{{280}} - \dfrac{1}{\infty }} \right)$
$\Rightarrow \dfrac{1}{{56}} = \left( {n - 1} \right)\left( {\dfrac{9}{{280}}} \right)$
Hence, by cross multiplying and $n$ subject, we have
$n = \dfrac{{280}}{{9 \times 56}} + 1$
$\Rightarrow n = \dfrac{{14}}{9}$
Hence, the correct option is B.

Note
For clarity, note the difference between the when the lens maker equation is used and when we simply say the radius of curvature is twice the focal length. The latter is used when the whole block is considered while the former is used for thin lenses or, as in this case, when a surface is considered just a surface.