Question

The object and the image are at a distance of 9cm and 16 cm respectively from the focal length of an equiconvex lens of radius curvature 12cm. The refractive index of the material of the lens is:
$$\begin{gathered} (a)\text{ 1}\text{.45} \\ (b)\text{ 1}\text{.5} \\ (c)\text{ 1}\text{.55} \\ (d)\text{ 1}\text{.6} \end{gathered}$$

Answer 1

- Hint: In the question consider R= 12cm, u = 9cm and v = 16cm that is the radius of curvature, object distance and the image distance respectively. Use the concept that the Focal length is the square root of the product of image distance and the object distance from the focal length that is $f=\sqrt{uv}$. Then use the lens maker formula that is ${\displaystyle \frac{1}{f}}=\left(\mu -1\right){\displaystyle \frac{2}{R}}$, this will help approaching the problem.

Complete step-by-step solution -

Given data:
Radius of curvature (R) of the equiconvex lens is 12 cm
Therefore, R =12cm.
The object and image distance from the focal length of an equiconvex lens are 9cm and 16 cm.
Therefore, u = 9cm and v = 16cm.
Where, u is the object distance from the focus and v is the image distance from the focus.
Now according to Newton’s formula the focal length of the lens in terms of the image and object distance from the focal length is given as
Focal length is the square root of the product of image distance and the object distance from the focal length.
$\Rightarrow f=\sqrt{uv}$cm
Now substitute the values we have,
$\Rightarrow f=\sqrt{9\times 16}=\sqrt{144}=12$cm.
Now according to the lens maker formula we have,
$\Rightarrow {\displaystyle \frac{1}{f}}=\left(\mu -1\right){\displaystyle \frac{2}{R}}$
Where,
f = focal length of the lens.
$$\mu$$ = refractive index of the lens.
R = radius of curvature of the lens.
Now substitute all the values in the above equation we have,
$\Rightarrow {\displaystyle \frac{1}{12}}=\left(\mu -1\right){\displaystyle \frac{2}{12}}$
Now simplify this we have,
$\Rightarrow 1=2\left(\mu -1\right)$
$\Rightarrow \mu -1={\displaystyle \frac{1}{2}}$
$\Rightarrow \mu =1+{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}=1.5$cm.
So the refractive index of the lens is 1.5cm.
So this is the required answer.
Hence option (B) is the correct answer.

Note – The trick point here was the lens maker formula, lens makers help establish the relationship between the radius of curvature, focal length and the refractive index of the material from which the lens is formed. An equiconvex lens is one in which the two opposite curved surfaces are related to each other via the same slope or the same shape.