Question

A 2.0 cm tall object is place perpendicular to the principal axis of a concave lens of focal length 15 cm. At what distance from the lens, should the object be placed so that it forms an image 10 cm from the lens? Also find the nature and the size of image formed.

Answer 1

Answer:

According to the question;

Object distance = u;

Image distance (v) = -10cm;

Focal length = -15cm;

By lens formula;

1/v-1/u = 1/f

(1/-10) -(1/u) = 1/-15

1/u = 1/15-1/10

1/u = -1/30

u = -30cm.

Thus, object should be placed 30cm in front of lens.

Now; Height of object h1= 2cm;

Magnification = h2/h1 = v/u

Putting values of v and u

Magnification =h2/5 = -10/-30

h2/5 = 1/3

h2 = 5/3

h2 = 1.67

Height of image is 1.67 cm.

Thus, the image is virtual, diminished, and erect and one-third of the size of object.

Answer 2

Object should be placed 30cm in front of lens.

Height of image is 0.67 cm.

The image is virtual, diminished, and erect.

Explanation:

Given object distance = u

Image distance v = -10cm

Focal length f = -15cm

Substituting the values in the lens formula, we get:

1/v -1/u = 1/f

(1/-10) - (1/u) = 1/-15

1/u = 1/15 - 1/10

1/u = (+2 -3) /30

 1/u =  -1/30

Therefore u = -30cm

So object should be placed 30cm in front of lens.

Given height of object h(o) = 2cm

Magnification = h(i) / h(o) = v/u

Substituting, we get:

h(i) / 2 = -10/-30

h(i) / 2 = 1/3

h(i) = 2/3

= 0.67 cm

Height of image is 0.67 cm.

Thus, the image is virtual, diminished, and erect.