Question

A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has to be formed to form the image of an object. At what distance from this lens, an object is to be placed in order to have a real image of the size of the object?
${\text{A}}{\text{.}}$ 20 cm
${\text{B}}{\text{.}}$ 30 cm
${\text{C}}{\text{.}}$ 60 cm
${\text{D}}{\text{.}}$ 80 cm

Answer 1

Hint- Here, we will proceed by determining the value of the focal length of the equivalent system (consisting of the plano-convex lens and the mirror). Then, we will use the concept that for magnification of 1, the object’s distance should be equal to the radius of curvature of the equivalent system.

Step By Step Answer:
Formulas Used-
$\dfrac{1}{{{{\text{F}}_{{\text{eq}}}}}} = 2\dfrac{1}{{{{\text{F}}_{\text{l}}}}} + \dfrac{1}{{{{\text{F}}_{\text{m}}}}}$, $\dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \left( {\dfrac{{{\mu _{{\text{inside}}}}}}{{{\mu _{{\text{outside}}}}}} - 1} \right)\left( {\dfrac{1}{{{{\text{R}}_1}}} - \dfrac{1}{{{{\text{R}}_2}}}} \right)$, ${{\text{F}}_{\text{m}}} = \dfrac{{\text{R}}}{2}$ and ${{\text{R}}_1} = \infty$ (for plane surface).

Let us consider a plano-convex lens having two regions I and II as shown in figure. The medium of regions I is air while that of region II is different (not air) with 1.5 as the refractive index. Let us consider the distance measured (from the pole of the lens or mirror) to the right as positive and those measured to the left as negative.

Given, Refractive index of the plano-convex lens (inside region II) ${\mu _{{\text{II}}}} = 1.5$
Radius of curvature of the plano-convex lens R = -30 cm

Here, a ray of light moves from region I towards region II. At the plane surface of the plano-convex lens, refraction occurs (due to the change of the medium) then at the curved surface of the plano-convex lens, reflection occurs (due to the silver polishing) and finally at the plane surface of the plano-convex lens, refraction of the reflected ray occurs (due to the change of the medium).

So, two refractions and one reflection occurs.

As we know that the focal length of the whole combination (where two refractions and one reflection occurs) of lens and mirror is given by
$\dfrac{1}{{{{\text{F}}_{{\text{eq}}}}}} = 2\dfrac{1}{{{{\text{F}}_{\text{l}}}}} + \dfrac{1}{{{{\text{F}}_{\text{m}}}}}{\text{ }} \to {\text{(1)}}$ where ${{\text{F}}_{\text{l}}}$ denotes the focal length of the lens, ${{\text{F}}_{\text{m}}}$ denotes the focal length of the mirror and ${{\text{F}}_{{\text{eq}}}}$ denotes the focal length of the equivalent system.

Also we know that the focal length for a convex lens is given by

$\dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \left( {\dfrac{{{\mu _{{\text{inside}}}}}}{{{\mu _{{\text{outside}}}}}} - 1} \right)\left( {\dfrac{1}{{{{\text{R}}_1}}} - \dfrac{1}{{{{\text{R}}_2}}}} \right){\text{ }} \to {\text{(2)}}$ where ${{\text{F}}_{\text{l}}}$ denotes the focal length of the lens, ${\mu _{{\text{inside}}}}$ denotes the refractive index of the material inside the lens, ${\mu _{{\text{outside}}}}$ denotes the refractive index of the material outside the lens, ${{\text{R}}_1}$ and ${{\text{R}}_2}$ denotes the radii of curvatures of the first and the second surface of the lens respectively.

For the given plano-convex lens, the first surface is plane surface with radius of curvature as ${{\text{R}}_1} = \infty$ and the second surface as convex surface with radius of curvature as R = -30 cm

Using the formula given in equation (2) for the plano-convex lens, we can write

$\dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \left( {\dfrac{{{\mu _{{\text{II}}}}}}{{{\mu _{\text{I}}}}} - 1} \right)\left( {\dfrac{1}{\infty } - \dfrac{1}{{\left( { - 30} \right)}}} \right)$

By substituting ${\mu _{{\text{II}}}} = 1.5$ and ${\mu _{\text{I}}} = 1$ (the refractive index of air is 1) in the above equation, we get

$\Rightarrow \dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \left( {\dfrac{{1.5}}{1} - 1} \right)\left( {0 + \dfrac{1}{{30}}} \right) \\\ \Rightarrow \dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{30}}} \right) \\\ \Rightarrow \dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \dfrac{{0.5}}{{30}} \\\ \Rightarrow \dfrac{1}{{{{\text{F}}_{\text{l}}}}} = \dfrac{1}{{60}} \\\ \Rightarrow {{\text{F}}_{\text{l}}} = 60{\text{ cm}} \\\$

Since, focal length of the mirror with radius of curvature as R is given by

${{\text{F}}_{\text{m}}} = \dfrac{{\text{R}}}{2}$

By substituting R = -30 cm in the above formula, we get

${{\text{F}}_{\text{m}}} = \dfrac{{{\text{30}}}}{2} = 15{\text{ cm}}$
In the formula given by equation (1) only positive values of the focal lengths will be taken. If any focal length is negative then, its modulus is taken.
By putting ${{\text{F}}_{\text{l}}} = 60{\text{ cm}}$ and ${{\text{F}}_{\text{m}}} = \left| { - 15} \right| = 15{\text{ cm}}$ in the formula given by equation (1), we get

$\Rightarrow \dfrac{1}{{{{\text{F}}_{{\text{eq}}}}}} = 2\left( {\dfrac{1}{{{\text{60}}}}} \right) + \dfrac{1}{{15}} \\\ \Rightarrow \dfrac{1}{{{{\text{F}}_{{\text{eq}}}}}} = \dfrac{1}{{{\text{30}}}} + \dfrac{1}{{15}} = \dfrac{{1 + 2}}{{30}} = \dfrac{3}{{30}} \\\ \Rightarrow \dfrac{1}{{{{\text{F}}_{{\text{eq}}}}}} = \dfrac{1}{{10}} \\\ \Rightarrow {{\text{F}}_{{\text{eq}}}} = 10{\text{ cm}} \\\$

Since, in order to have a real image of the size of the object, the object should be placed at the centre of curvature of the equivalent system (magnification is equal to 1 and object’s distance is equal to image’s distance). This means that the distance of the object from the lens should be equal to the radius of curvature of the equivalent system.

i.e., Required distance of the object from the lens = ${{\text{R}}_{{\text{eq}}}} = 2{{\text{F}}_{{\text{eq}}}} = 2 \times 10 = 20{\text{ cm}}$

Therefore, option A is correct.

Note: In this particular problem, we did not consider the region which is to the right of the convex surface of the given plano-convex lens because the curved surface of the given plano-convex lens is silvered polished. Also, we have taken R = -30 cm because the radius of the curvature is measured to the left from the pole of the lens.