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 been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object?

• A. 20 cm
• B. 30 cm
• C. 60 cm
• D. 80 cm

Answer 1

Answer: (A)

A plano-convex lens behaves as a concave mirror if its one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface.
Therefore, in this lens two refractions and one reflection occur.
Let the focal length of silvered lens is $F$.
$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}+\frac{1}{f_{m}}$
$=\frac{2}{f}+\frac{1}{f_{m}}$
where $f=$ focal length of lens before silvering
$f_{m}=$ focal length of spherical mirror
$\therefore \frac{1}{F}=\frac{2}{f}+\frac{2}{R}\,\,\,...(i)$
$\left(\because R=2 f_{m}\right)$
Now, $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\,\,\,...(ii)$
Here, $R_{1}=\infty, R_{2}=30\, cm$
$\therefore \frac{1}{f}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{30}\right)$
$\Rightarrow \frac{1}{f}=-\frac{0.5}{30}=-\frac{1}{60}$
$\Rightarrow f=-60\, cm$
Hence, from Eq. (i) $\frac{1}{F}=\frac{2}{60}+\frac{2}{30}=\frac{6}{60}$
$F=10 \,cm$
Again given that,
size of object $=$ size of image
$\Rightarrow O=I$
$\therefore m=-\frac{v}{u}=\frac{I}{O}$
$\Rightarrow \frac{v}{u}=-1$
$\Rightarrow v=-u$
Thus, from lens formula
$\frac{1}{F}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{10}=\frac{1}{-u}-\frac{1}{u}$
$\Rightarrow \frac{1}{10}=-\frac{2}{u}$
$\therefore u=-20 \,cm$
Hence, to get a real image, object must be placed at a distance $20\,cm$ on the left side of lens.