Question

Figure shows a lens of refractive index $\mu =1.5$. $C_1$ and $C_2$ are the centres of curvature of the two faces of the lens of radii of curvature $10\text{cm}$ and $15\text{cm}$ respectively. The lens behaves as a

• A. converging lens of focal length $60\text{cm}$
• B. converging lens of focal length $12\text{cm}$
• C. diverging lens of focal length $60\text{cm}$
• D. diverging lens of focal length $12\text{cm}$

Answer 1

The correct option is A converging lens of focal length $60\text{cm}$


From the data given in the question,
$R_1=+10\text{cm}$ and $R_2=+15\text{cm}$

Using lens maker's formula,

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

Substituting the data given in the question we get,

$\frac{1}{f}=(1.5-1)(\frac{1}{10}-\frac{1}{15})$

$\Rightarrow f=+60\text{cm}$,

Positive sign shows it is a converging lens.

Hence, option (a) is the correct answer.