Question

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

• A. greater than zero but less than 1.5
• B. greater than 1.5 but less than 2.0
• C. greater than 2.0 but less than 2.5
• D. greater than 2.5 but less than 2.0

Answer 1

Answer: (A)

Focal length of lens,
$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For equi-convex lens,
$R_1=+R,R_2=-R$
$\therefore \frac{1}{f}=(\mu -1)\left(\frac{1}{R}-\frac{1}{-R}\right)$
$\frac{1}{f}=(\mu -1)\left(\frac{2}{R}\right)$
$f=\frac{R}{2(\mu -1)}$
$f<R,$
so, $2(\mu -1)<1$
i.e. $(\mu -1)<\frac{1}{2}$
$(\mu -1)<0.5$
$\mu <1.5$
Focal length of convex lens is positive. So, $\mu$ cannot be negative, hence $\mu$ should be greater than zero but less than $1.5$