Question

The radii of curvature of the two surfaces of a lens are $20\text{cm}$ and $30\text{cm}$ and the refractive index of the material of the lens is $1.5$. If the lens is concavo convex then , the focal length of the lens is $($ in $\text{cm})$

Answer 1

For concavo convex lens ;

$R_1=+20\text{cm},R_2=+30\text{cm}$ and $\mu =1.5$

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,

$\frac{1}{f}=(1.5-1)(\frac{1}{20}-\frac{1}{30})$

$\Rightarrow$$\frac{1}{f}=0.5\times \left(\frac{1}{60}\right)$

$\therefore f=\frac{600}{5}=120\text{cm}$