Question

The object and the image are at distance of $9cm$ and $16cm$ respectively from the focal length of an equiconvex lens of radius of curvature $12cm$. The refractive index of the material of the lens is:

• A. 1$.45$
• B. $1.5$
• C. $1.55$
• D. $1.6$

Answer 1

Answer: (B)

$1.5$

Given: $u=9cm$ and $v=16cm$

Using Newton's formula, we can write the focal length of the lens as:

$f=\sqrt{uv}$

$=\sqrt{9\times 16}=12cm$

Now, applying Lens maker's formula, we have

$\Rightarrow \frac{1}{f}=(\mu -1)\frac{2}{R}$

$\Rightarrow \frac{1}{12}=(\mu -1)\frac{2}{12}$ ....($\because R=12cm$)

$\Rightarrow \frac{1}{12}=(\mu -1)\frac{1}{6}$

$\Rightarrow \frac{1}{2}=(\mu -1)$

$\Rightarrow \mu =1+\frac{1}{2}=1.5$