Question

A Plano convex lens, when silvered in the plane side, behaves like a concave mirror of focal length 30 cm. however, when silvered on the convex side it behaves like a concave mirror of focal length 10 cm. then the refractive index of its material will be
A. 3.0
B. 2.0
C. 2.5
D. 1.5

Answer 1

Hint: In this question we have to find the refractive index of the material of the lens. For this use the formula of focal length, and then use the relation of focal length with radius of curvature and refractive index.

Complete step by step answer:
Given,
When silvered in the plane side the Plano convex lens behaves like a concave mirror of focal length 30 cm.
When silvered on the convex side the Plano convex lens behaves like a concave mirror of focal length 10 cm.
Now we will use formula of focal length to find focal length of the lens
$\dfrac{1}{F} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_m}}}$
${f_m}$ is the focal length of the mirror in cm,
${f_1}$ is the focal length in cm
F is focal length of lens in cm
$\dfrac{1}{F} = \dfrac{2}{{{f_1}}} + \dfrac{1}{{{f_m}}}$……. (1)
As we know that the focal length of mirror will be infinity, so ${f_m}$=0
The relation between focal length, radius of curvature and refractive index is given below,
$\dfrac{1}{{{f_1}}} = (\mu - 1)\left( {\dfrac{1}{R}} \right)$
Where,
$\mu $is the refractive index
R is the radius of curvature cm.
Putting the values of F, ${f_1}$and ${f_m}$in equation (1)
$\dfrac{1}{{30}} = 2(\mu - 1)\left( {\dfrac{1}{R}} \right) + \dfrac{1}{\infty }$
$\dfrac{1}{{30}} = 2(\mu - 1)\left( {\dfrac{1}{R}} \right) + 0$
$\dfrac{1}{{30}} = 2(\mu - 1)\left( {\dfrac{1}{R}} \right)$
$30 = \dfrac{R}{{2(\mu - 1)}}$
$60 = \dfrac{R}{{(\mu - 1)}}$…….. (2)
Now, using the formula of focal length when silvered on the convex side. Let focal length is ${F_1}$
$\dfrac{1}{{{F_1}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_m}}}$
In this ${f_m}$case will be equal to $\dfrac{R}{2}$
\[\dfrac{1}{{{F_1}}} = \dfrac{2}{{{f_1}}} + \dfrac{1}{{R/2}}\]
\[\dfrac{1}{{{F_1}}} = 2(\mu - 1)\left( {\dfrac{1}{R}} \right) + \dfrac{1}{{R/2}}\]
\[\dfrac{1}{{{F_1}}} = 2(\mu - 1)\left( {\dfrac{1}{R}} \right) + \dfrac{2}{R}\]
\[\dfrac{1}{{10}} = 2\left( {\dfrac{1}{{60}}} \right) + \dfrac{2}{R}\]
\[\dfrac{1}{{10}} = \dfrac{1}{{30}} + \dfrac{2}{R}\]
\[\dfrac{1}{{10}} - \dfrac{1}{{30}} = \dfrac{2}{R}\]
\[\dfrac{2}{{30}} = \dfrac{2}{R}\]
\[R = 30{\text{ }}cm\]
Putting the value of R in equation (2)
\[60 = \dfrac{{30}}{{(\mu - 1)}}\]
\[(\mu - 1) = \dfrac{{30}}{{60}}\]
\[(\mu - 1) = \dfrac{1}{2}\]
\[\mu = \dfrac{1}{2} + 1\]
\[\mu = \dfrac{3}{2}\]
$\therefore$ the value of the refractive index of lens is, \[\mu = 1.5\]

Hence, the correct answer is option (D).

Note: From the above we found the refractive index of a Plano convex lens. As calculated above just use the formula of focal length of the lens. But remember that the focal length of parts of the lens when will be infinity and when will be equal to half of the radius of curvature.