Question

A double convex lens made of glass of refractive index 1.5 has both radii of curvature of 20cm. An object 2cm high is placed at 10cm from the lens. Find the position, nature and size of the image.

Answer 1

Hint: As a first step, one could read the question carefully and hence note down the given values as per the sign convention. Then you could recall the lens maker's formula, using which you would find the focal length of the lens which you could further substitute in the lens formula to get the image distance. In order to get the nature of the image formed, one could use the formula for magnification.
Formula used:
Lens maker’s formula,
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
Lens formula,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Complete answer:
In the question we have a double convex/biconvex lens of same radii of curvature 20cm, that is,
${{R}_{1}}={{R}_{2}}=20cm$
The object distance is given to be 10cm from the lens to the left, that is,
$u=-10cm$
Now we have the height of the object given as,
${{h}_{o}}=2cm$
Let us assume that the incident ray comes from air with refractive index 1 on to glass of refractive index 1.5. Now let us recall lens maker’s formula which is given by,
$\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
Now let us substitute the given values by following sign convention to get,
$\dfrac{1}{f}=\left( 1.5-1 \right)\left( \dfrac{1}{20}+\dfrac{1}{20} \right)=\dfrac{1}{20}$
$\Rightarrow f=20cm$
Now let us recall the lens formula,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Now let us substitute the given values to get,
$\dfrac{1}{20}=\dfrac{1}{v}-\dfrac{1}{-10}=\dfrac{1}{v}+\dfrac{1}{10}$
$\Rightarrow \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{-10}{200}$
$\therefore v=-20cm$
So we found that the image is formed on the left side of the mirror, hence it is a virtual image. Now from the expression for magnification we have,
$m=\dfrac{-{{h}_{i}}}{{{h}_{o}}}=\dfrac{-v}{u}$
$\Rightarrow {{h}_{i}}=\dfrac{v}{u}\times {{h}_{o}}=\dfrac{-20}{-10}\times 2$
$\therefore {{h}_{i}}=4cm$
Therefore, we found that the image formed of the object is erect.

Note:
Let us discuss the sign conventions which played a very important role in the above solution. All the measurements made to the left of the optic centre are taken negative and those to the right are taken positive. And the measurements above the principal axis are positive and those below are negative.