Question

Find the image distance given that the object distance is $6.5cm$ a thin lens with two radii$r_1=+10$ and $r_2=-10$ and the index of refraction is $1.5$. Also the object is on the left side closest to $r_1$.

Answer 1

Hint : In order to solve the given question we need to know the lens formula and its application. Also we need to have an idea of application of lens maker’s formula. First of all by using the lens maker’s formula we need to find the focal length of the lens. Then by using the simple lens formula we need to find the image distance. Solving the equation formed will give us the solution of the given question.

Complete step by step solution:
Step one
The value of radius is given in question as,$r_1=10cm$and$r_2=-10cm$
Also the value of refractive index is given as,$\mu =1.5$
The object is placed at a distance of,$u=6.5cm$
Step two
Now, we need to write the lens maker’s formula
${\displaystyle \frac{1}{f}}=(\mu -1)\left({\displaystyle \frac{1}{r_1}}-{\displaystyle \frac{1}{r_2}}\right)$…………..(i)
Now we need to put the values in equation (i).
After putting the values in equation (i), we get
${\displaystyle \frac{1}{f}}=(1.5-1)\left({\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{10}}\right)$
$\Rightarrow {\displaystyle \frac{1}{f}}=0.5\times {\displaystyle \frac{1}{5}}={\displaystyle \frac{1}{10}}$
$\therefore f=10cm$
Step three
Now in order to find the image distance we need to use the lens formula.
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$………………. (ii)
After putting the values in equation (ii), we get,
${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{6.5}}={\displaystyle \frac{1}{10}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{6.5}}$
$\Rightarrow {\displaystyle \frac{1}{v}}={\displaystyle \frac{33}{130}}$
$\therefore v={\displaystyle \frac{130}{33}}=3.94cm$

Therefore, the image is formed at a distance of $3.94cm$

Note: The lens maker’s formula is used by the lens manufacturers to calculate the power of the combination of lenses of a given refractive index. The lens maker’s formula gives the relation between the focal length of a lens and the refractive index of the lens’s material and the radii of curvature of the lenses. The focal length of a lens depends on the radii of curvature of the lens and refractive index of the lens.