An object placed at a distance 30 cm in front of a lens produces clear inverted image at a distance 60 cm from the lens. If the object is placed at 60 cm from the lens, then it produces a clear inverted image at a distance of …………… from the lens.

1. 20 cm
2. 30 cm
3. 60 cm
4. 90 cm

30 cm

Reason

Case 1 :

Given,

• Object distance (u) = -30 cm
• Image distance (v) = +60 cm

Let, focal length be 'f'.

From lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1}{60}-\dfrac{1}{(-30)} \\[1em] \Rightarrow \dfrac{1}{\text f} =\dfrac{1}{60}+\dfrac{1}{30} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1+2}{60}\\[1em] \dfrac{1}{\text f} = \dfrac{3}{60} \\[1em] \Rightarrow \text f = \dfrac{60}{3} = +20 \text { cm}$

Case 2 :

Given,

• Object distance (u) = -60 cm
• Focal length (f) = +20 cm

Let, image distance be 'v'.

Again using lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1em] \Rightarrow \dfrac{1}{\text v} = \dfrac{1}{\text f} + \dfrac{1}{\text u} \\[1em] \Rightarrow \dfrac{1}{\text v} = \dfrac{1}{20}+\dfrac{1}{(-60)} \\[1em] \Rightarrow \dfrac{1}{\text v} =\dfrac{1}{20}-\dfrac{1}{60} \\[1em] \Rightarrow \dfrac{1}{\text v} = \dfrac{3-1}{60}\\[1em] \Rightarrow \dfrac{1}{\text v} = \dfrac{2}{60} \\[1em] \Rightarrow \text v = \dfrac{60}{2} = +30 \text { cm}$