Question 11: Ray Optics

Question

A thin biconvex lens is prepared from the glass $(\mu=1.5)$ both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of $\_\_\_\_$ cm.

Accepted Answer

When one surface of a lens is silvered, it behaves like a mirror. The effective system consists of a lens, followed by a mirror, and then the lens again (as light passes through the lens, reflects at the silvered surface, and passes back through the lens). The effective power ( $\mathrm{P}_{\mathrm{eq}}$ ) of this silvered lens is given by:
$$ \mathrm{P}_{\mathrm{eq}}=2 \mathrm{P}_{\mathrm{l}}+\mathrm{P}_{\mathrm{m}} $$
Where, $\mathrm{P}_{\mathrm{l}}=$ Power of the lens and $\mathrm{f}_{\mathrm{m}}=$ Focal length of the silvered surface (mirror).
The power of lens is given as $P=\frac{1}{f}$, where focal length (f) should be in meters.
The power of mirror is given as $P=-\frac{1}{f}$, where focal length (f) should be in meters.
Using the Lens Maker's Formula:
$$ \frac{1}{\mathrm{f}_{\mathrm{l}}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}ight) $$
For a biconvex lens, the refractive index of material of lens is $\mu=1.5$, the radius of curvature for the first surface is $\mathrm{R}_1=+20 \mathrm{~cm}$ (convex surface) and $\mathrm{R}_2=-20 \mathrm{~cm}$ is for the second surface (opposite convex surface)
$$ \frac{1}{f_1}=(1.5-1)\left(\frac{1}{20}-\left(\frac{1}{-20}ight)ight) $$
$$\Rightarrow $$ $$ \frac{1}{\mathrm{f}_1}=0.5\left(\frac{1}{20}+\frac{1}{20}ight)=0.5 	imes \frac{2}{20}=\frac{1}{20} $$
So, the focal length of lens is $\mathrm{f}_{\mathrm{l}}=20 \mathrm{~cm}$.
The power of lens is, $\mathrm{P}_{\mathrm{l}}=\frac{1}{(20 / 100)}=5 \mathrm{D}$.
Since the silvered surface is the back surface of a biconvex lens, it acts as a concave mirror for the light inside the lens.
So, the radius of curvature is $\mathrm{R}=-20 \mathrm{~cm}$.
$$ \mathrm{f}_{\mathrm{m}}=\frac{\mathrm{R}}{2}=-\frac{20}{2}=-10 \mathrm{~cm} $$
The optical power of mirror is given as $\mathrm{P}_{\mathrm{m}}=-\frac{1}{\mathrm{f}_{\mathrm{m}}}$, where focal length should be in meters.
The power of mirror is, $P_m=-\frac{1}{(-10 / 100)}=10 \mathrm{D}$.
Using the effective power formula,
$$ \mathrm{P}_{\mathrm{eq}}=(2 	imes 5+10) \mathrm{D}=20 \mathrm{D} $$
As the effective focal length of the system is +20 D , and the whole system behaves like a mirror,
$$ f_{\mathrm{eq}}=-\frac{1}{P_{\mathrm{eq}}} \Rightarrow f_{\mathrm{eq}}=-\frac{1}{20} \mathrm{~m}=-\frac{100}{5} \mathrm{~cm}=-5 \mathrm{~cm} $$
When the object is placed at the centre of curvature ( $C$ ), the position of image and object are at the same place. So, the object distance is equal to the radius of curvature.
$$ \mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{R}_{\mathrm{eq}}}{2} \Rightarrow \mathrm{R}_{\mathrm{eq}}=2 	imes \mathrm{f}_{\mathrm{eq}}=2 	imes(-5) \mathrm{cm} $$
$$\Rightarrow $$ $$ \mathrm{R}_{\mathrm{eq}}=-10 \mathrm{~cm} $$
Therefore, to have the image and object at the same place, the object should be placed at 10 cm .
Hence, the correct option is (A).

Answer

10

Answer

12.5

Answer

13

Answer

13.5

Ray Optics

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