Question

(a) Two lenses have power of (i) + 2 D (ii) −4 D. What is the nature and focal length of each lens?
​(b) An object is kept at a distance of 100 cm from each of the above lenses. Calculate the (i) image distance (ii) magnification in each of the two cases.

Answer 1

(a) Power of a lens is given by reciprocal of its focal length. i.e. $P=\frac{1}{f(\text{in}\text{meter)}}$

A negative power means a negative focal length and positive power means positive focal length. As we know, a converging or convex lens has a positive focal length and a diverging or concave lens has a negative focal length. Therefore,
(i) The lens with positive power '+2D' is a converging lens
(ii) The lens with negative power '−4 D' is a diverging lens in nature.

(b) (i) Power of the lens = +2D, $P=\frac{1}{f(\text{in}\text{meter)}}$

Focal length $=\frac{1}{P}=\frac{1}{2}\text{m}=50\text{cm}$

Object distance = $-$100 cm

Using the lens formula,

$$\begin{gathered} \frac{1}{v}-\frac{1}{(-100)}=\frac{1}{50} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=\frac{{\displaystyle 1}}{{\displaystyle 100}} \\ \Rightarrow v=100\text{cm} \end{gathered}$$

So, the image distance for the first lens is 100 cm.

Magnification of the lens = $\frac{v}{u}=\frac{100\text{cm}}{-100\text{cm}}=-1$

(ii) Power of the lens = -4D, $P=\frac{1}{f(\text{in}\text{meter)}}$

Focal length $=\frac{1}{P}=-\frac{1}{4}\text{m}=-25\text{cm}$

Object distance = $-$100 cm

Using the lens formula,

$$\begin{gathered} \frac{1}{v}-\frac{1}{(-100)}=-\frac{1}{25} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=-\frac{{\displaystyle 1}}{{\displaystyle 20}} \\ \Rightarrow v=-20\text{cm} \end{gathered}$$

So, the image distance for the second lens is $-$25 cm.

Magnification of the lens = $\frac{v}{u}=\frac{-25\text{cm}}{-100\text{cm}}=\frac{1}{4}=0.25$