Question

The radius of curvature for a convex lens is 40 cm each surface its refractive index is 1.5 the focal length will be

Answer 1

Let the radius of curvature of the convex lens are denoted as $R_{1}\ and\ R_{2}\ respectively$

For a convex lens $R_{1} =+40\ cm$  [As it is towards right]

                              $R_{2} =\ -40\ cm$ [As it is towards left]

The refractive index of the lens  $[\mu]=1.5$

Let f is the focal length of the lens.

From lens maker's formula we know that -

                                $\frac{1}{f} =[\mu -1][\frac{1}{R_{1}}- \frac{1}{ R_{2}}]$

                                $\frac{1}{f} =[1.5-1][\frac{1}{40}- \frac{1}{(-40)} ]$

                                $\frac{1}{f} =0.5*\frac{2}{40}$

                                $\frac{1}{f} =\frac{1}{40}$

                                $f=40\ cm$      [ans]

Hence, the focal length of the lens is 40 cm.

Answer 2

Hello friend,

For any convex lens there are two surfaces i.e. left and right

Given -

For right surface R1 = 40cm = 0.4m

For left surface R2 = -40cm = -0.4m

µ = 1.5

f = ?

Solution -

By lens makers formula,

1/f = (µ-1) (1/R1 - 1/R2)

1/f = (1.5-1)(1/0.4 - 1/-0.4)

1/f = 0.5 × (2/0.4)

f = 0.4m = 40 cm

Focal length of lens is 40 cm.

Hope I helped...