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.
- (a)20 cm
- (b)30 cm
- (c)60 cm
- (d)90 cm
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30 cm
Reason
Case 1 :
Given,
- Object distance (u) = -30 cm
- Image distance (v) = +60 cm
Let, focal length be 'f'.
From lens formula,
$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1em] \Rightarrow \frac{1}{\text f} = \frac{1}{60}-\frac{1}{(-30)} \\[1em] \Rightarrow \frac{1}{\text f} =\frac{1}{60}+\frac{1}{30} \\[1em] \Rightarrow \frac{1}{\text f} = \frac{1+2}{60}\\[1em] \frac{1}{\text f} = \frac{3}{60} \\[1em] \Rightarrow \text f = \frac{60}{3} = +20 \text { cm}\end{aligned}$$
Case 2 :
Given,
- Object distance (u) = -60 cm
- Focal length (f) = +20 cm
Let, image distance be 'v'.
Again using lens formula,
$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1em] \Rightarrow \frac{1}{\text v} = \frac{1}{\text f} + \frac{1}{\text u} \\[1em] \Rightarrow \frac{1}{\text v} = \frac{1}{20}+\frac{1}{(-60)} \\[1em] \Rightarrow \frac{1}{\text v} =\frac{1}{20}-\frac{1}{60} \\[1em] \Rightarrow \frac{1}{\text v} = \frac{3-1}{60}\\[1em] \Rightarrow \frac{1}{\text v} = \frac{2}{60} \\[1em] \Rightarrow \text v = \frac{60}{2} = +30 \text { cm}\end{aligned}$$