A convex lens has focal length 12 cm with an object at a distance of 20 cm in front of the lens. He obtains a blurred image on the screen placed at a distance of 23 cm in front of the lens. In order to obtain the clear image, he has to move the screen

1. towards the lens.
2. away from the lens.
3. to a position very far away from the lens.
4. either towards or away from the lens.

away from the lens.

Reason —

Given,

Focal length (f) = +12 cm

Object distance (u) = -20 cm

Blurred image distance (v') = +23 cm

Let, clear image distance be v.

From lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} – \dfrac{1}{\text u} \\[1 em] \dfrac{1}{\text v} = \dfrac{1}{\text f} + \dfrac{1}{\text u} \\[1 em] \dfrac{1}{\text v}=\dfrac{1}{12} + \dfrac{1}{(-20)}\\[1 em] \dfrac{1}{\text v}=\dfrac{1}{12} - \dfrac{1}{20}\\[1 em] \dfrac{1}{\text v}=\dfrac{5-3}{60 }= \dfrac{2}{60}\\[1 em] \text v = \dfrac{60}{2} = +30 \text { cm}$

Now,

Clear image distance (v) - Blurred image distance (v') = 30 - 23 = 7 cm

So, in order to obtain the clear image, he has to move the screen 7 cm away from the lens.