Question

an object 3 cm high is placed perpendicular to the principal axis of a concave lens of focal length 7.5CM the image is formed at a distance of 5 cm from the lens calculate distance at which object is placed and find size and nature of image formed

Answer 1

Using lens formula 1/v-1/u = 1/f

We have 1/u = 1/v-1/f = 1/-5-1/-7.5 = -3+2/15 = -1/15

u = -15 cm

The object should be placed 15 cm from the concave lens,

m = v/u = Image size/Object size

Image size = v/u x Object size

I = -5/-15 x 3 = 1 cm

The image is virtual and erect and has a size 1 cm.

Answer 2

Answer:

Explanation:

h1=3.0cm,f=−7.5cm,v=−5.0cm,v=?,h2=?

From 1f=1v−1u,1u=1v−1f=1−5−1−7.5=−3+215 or u=−15cm

i.e., object is at 15cm from the concave lens.

From h2h1=vu,h2=vu×h1=−5.0−15.0×3.0=1cm

The image is virtual and erect, and its size is 1cm