Question

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

Answer 1

(a) Power of lens A = +2D
Power of lens B = -4D

Power = $\frac{1}{\mathrm{focal}\mathrm{length}}$

or Focal length = $\frac{1}{\mathrm{power}}$

Thus, focal length (f_A ) of A = $\frac{1}{2}=0.5\mathrm{m}=50\mathrm{cm}$

Focal length (f_B ) of B = $\frac{1}{-4}=-0.25\mathrm{m}=-25\mathrm{cm}$

Therefore, lens A is a convex lens as it has a positive focal length and lens B is concave as it has a negative focal length.

(b)
Object distance (u) = -100 cm (sign convention)
For lens A:
Image distance (v) = ?

Using the lens formula, we get:

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}-\frac{1}{-100}=\frac{1}{50} \\ \Rightarrow \frac{1}{v}+\frac{1}{100}=\frac{1}{50} \\ \Rightarrow \frac{1}{v}=\frac{1}{50}-\frac{1}{100} \\ \Rightarrow \frac{1}{v}=\frac{2-1}{100} \\ \Rightarrow \frac{1}{v}=\frac{1}{100} \\ \therefore v=100\mathrm{cm}. \end{gathered}$$

∴Magnification = $\frac{v}{u}=\frac{100}{-100}=-1$.
For lens B:
Image distance (v₂) =?

Using the lens formula, we get:

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}-\frac{1}{-100}=\frac{1}{-25} \\ \Rightarrow \frac{1}{v}+\frac{1}{100}=\frac{-1}{25} \\ \Rightarrow \frac{1}{v}=-\frac{1}{25}-\frac{1}{100} \\ \Rightarrow \frac{1}{v}=\frac{-4-1}{100} \\ \Rightarrow \frac{1}{v}=\frac{-5}{100} \\ \therefore v=-20\mathrm{cm}. \end{gathered}$$

Negative sign shows that the image formed is virtual.

∴ Magnification = $\frac{\mathit{v}}{\mathit{u}}=\frac{-20}{-100}=-0.2$.