Question

A plano convex lens of refractive index 1.5 and radius of curvature 30cm is silvered on its flat surface. Where an object should be placed relative to the silvered lens so as to form a real image of the same size as the object?
A. 10 cm
B. 20 cm
C. 30 cm
D. 60 cm

Answer 1

The power of the lens is reciprocal of the focal length. Use the lens maker’s formula to express the focal length of the convex lens. Use the lens equation to determine the object distance for the size of the image to be equal to the size of the object.

Formula used:
Lens maker’s formula, $\dfrac{1}{{{f_c}}} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)$
Here, ${R_1}$ and ${R_2}$ are the radii of curvatures of the convex lens.

Complete step by step answer:
Let us determine the power of the combined lens using the formula,
${P_{com}} = {P_C} + {P_f}$
Here, ${P_C}$ is the power of the convex lens and ${P_f}$ is the power of the flat mirror.
Since the flat surface of the mirror is silvered, its power will be the zero. Therefore, we can write,
${P_{com}} = {P_C}$
We know that the power of the lens is the reciprocal of the focal length. Therefore,
$\dfrac{1}{{{f_{com}}}} = \dfrac{1}{{{f_C}}}$ …… (1)
Here, ${f_C}$ is the focal length of the convex lens.
We have the formula for the focal length of the lens of refractive index $\mu$,$\dfrac{1}{{{f_c}}} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)$
Here, ${R_1}$ and ${R_2}$ are the radii of curvatures of the convex lens.

Since the convex lens has symmetric surfaces on both sides, the radii of curvatures of the two surfaces are the same. Therefore,
$\dfrac{1}{{{f_c}}} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_1}}}} \right)$
$\Rightarrow \dfrac{1}{{{f_c}}} = \left( {\mu - 1} \right)\dfrac{2}{{{R_1}}}$
Using the above equation in equation (1), we get,
$\dfrac{1}{{{f_{com}}}} = \left( {\mu - 1} \right)\dfrac{2}{{{R_1}}}$
$\Rightarrow {f_{com}} = \dfrac{{{R_1}}}{{2\left( {\mu - 1} \right)}}$
Substituting ${R_1} = 30\,{\text{cm}}$ and $\mu = 1.5$ in the above equation, we get,
${f_{com}} = \dfrac{{30}}{{2\left( {1.5 - 1} \right)}}$
$\Rightarrow {f_{com}} = \dfrac{{30}}{1}$
$\Rightarrow {f_{com}} = 30\,{\text{cm}}$

For the size of the image to be equal to the size of the object, the image distance must be equal to the object distance. For the real image to produce, the image distance must be the negative. Using the lens equation,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Here, v is the object distance and u is the image distance.
Substituting $u = - v$ in the above equation, we get,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{{ - v}}$
$\Rightarrow f = \dfrac{v}{2}$
$\Rightarrow v = 2f$
Substituting $f = 30\,{\text{cm}}$ in the above equation, we get,
$v = 2\left( {30} \right)$
$\therefore v = 60\,{\text{cm}}$
Therefore, the object should be placed at 60 cm from the lens.

So, the correct answer is option D.

Note: Students must know that when the convex lens forms a real image and a virtual image. If the image forms on the same side of the object, the real image of the object is formed. If the image forms on the same side of the object, the image distance is taken as negative. Note that when the size of the image is the same as the size of the object, then the magnification of the lens is 1.