Question

An object is 2m from a lens which forms an erect image 1/4th the size of the object. Determine the focal length of the lens,what type of lens is this​

Answer 1

Distance of the object , u = - 2 m

Magnification , m = 1/4

let distance of image be v m

and focal length be f m

As we know ,

m = v/u

1/4 = v/2 (substitute the value of u & m)

so, v = -2 /4 v =-1/2 ,v = - 0.5 m

Lens formula,

1/f = 1/v-1/u

1/f = -1/0.5 -(-1/2)

= -1/0.5+1/2

= (-2+0.5)/0.5 × 2 (LCM of 0.5 and 2)

= -1.5/1

1/f = -1.5 /1

f = -1/1.5

f = -0.67 m

The image is small , erect, Therefore , it is a concave len.

Answer 2

Answer:

Explanation:

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

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