Question

A converging becam of light forms a sharp image on a screen. A lens is placed in the path of the beam, at $10\text{cm}$ from the screen, it is found that the screen has to be moved $8\text{cm}$ further away from the lens to obtam a sharp image. The focal length and the nature of lens is.

• A. $18.5\text{cm};\text{convex lens}$
• B. $20.5\text{cm};\text{concave lens}$
• C. $22.5\text{cm};\text{concave lens}$
• D. $19.5\text{cm};\text{convex lens}$

Answer 1

The correct option is C $22.5\text{cm};\text{concave lens}$

Let us consider the virtual object at $10\text{cm}$.
$u=10\text{cm}$
The final image at a distance $v=10+8=18\text{cm}.$
Lens formula
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{18}-\frac{1}{10}=\frac{1}{f}$
$\frac{1}{f}=\frac{10-18}{180}=\frac{-8}{180}$
$f=-\frac{180}{8}=-22.5\text{cm}$
$\therefore$ Used lens is concave in nature