Question

An object is 2 cm from a lens which forms an erect image 1/4th (exactly) the size of the object .Determine the focal length of the lens.
What type of lens is this?

Answer 1

Diminished erect image is formed in case of concave lens. So it is a concave lens.
object distance (u) = -2cm
image distance  = v cm
magnification (m) = +(1/4)

m = v/u
⇒ +1/4 = v/(-2)
⇒ v = -2/4 = -0.5 cm

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\ \\ \frac{1}{-2} + \frac{1}{-0.5} = \frac{1}{f} \\ \\ -0.5-2= \frac{1}{f} \\ \\ \frac{1}{f} =-2.5 \\ \\ f = \frac{1}{-2.5}=-0.4cm$

Focal length is -0.4 cm. It is a concave lens as indicated by the negative sign of focal length.

Answer 2

Answer:object distance (u) = -2cm

image distance  = v cm

magnification (m) = +(1/4)

m = v/u

⇒ +1/4 = v/(-2)

⇒ v = -2/4 = -0.5 cm

Focal length is -0.4 cm. It is a

concave lens.