Question

The focal length of a bi-convex lens is $20\text{cm}$ and its refractive index is $1.5$. If the radii of curvatures of two surfaces of lens are in the ratio $1:2$, then the value of the larger radius of curvature is:
$\left(\text{in cm}\right)$

• A. $10$
• B. $15$
• C. $20$
• D. $30$

Answer 1

The correct option is D $30$

From the data given in the question,

$\mu =1.5;f=20;R_1:R_2=1:2$

Let $R_1=R\left(\text{say}\right)$ then, $R_2=2R$

Using Lens maker's formula ,

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

For a bi-convex lens, $R_1$ is positive and $R_2$ is negative.

$\therefore \frac{1}{f}=(\mu -1)(\frac{1}{R_1}+\frac{1}{R_2})$

Substituting the given data we get,

$\frac{1}{20}=(1.5-1)(\frac{1}{R}+\frac{1}{2R})$

$\Rightarrow \frac{1}{20}=\frac{3}{4R}\Rightarrow$$R=15\text{cm}$

$\therefore R_2=2R=30\text{cm}$

Hence, option (d) is the correct answer.