Question

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. The position of the image is:

• A. -3 cm
• B. 3 cm
• C. -0.333 cm
• D. 0.333 cm

Answer 1

The correct option is A -3 cm

object distance, u = -4 cm focal length, f = -12 cm so, if $\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$ then

$\frac{1}{v}$ = $\frac{1}{f}$ + $\frac{1}{u}$

$\frac{1}{v}$ = $\frac{-1}{4}$ - $\frac{1}{12}⟹\frac{1}{v}=-\frac{1}{3}$

Thus, the image distance will be, v = -3 cm and it will be erect, diminished and virtual.