Question

A double convex lens has two surfaces of equal radii $15\text{cm}$ and refractive index $\mu =1.5$, its focal length is equal to

• A. $-15\text{cm}$
• B. $-30\text{cm}$
• C. $+15\text{cm}$
• D. $-30\text{cm}$

Answer 1

The correct option is C $+15\text{cm}$

Sign convention: Direction of incident ray taken as +ve.
Applying lens maker formula,
$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$


For the given double convex thin lens,
$R_1=+R=+15\text{cm},R_2=-15\text{cm}$
${\mu }_2=\mu =1.5,{\mu }_1=1$
Substituting the values in above formula
$\Rightarrow \frac{1}{f}=(\frac{1.5}{1}-1)(\frac{1}{15}-\frac{1}{(-15)})$
or $\frac{1}{f}=\left(0.5\right)\left(\frac{2}{15}\right)$
$\therefore f=15\text{cm}$
$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Be cautions in taking sign of}{R}_1\text{and}{R}_2\text{while using lens maker's formula}\end{array}$$