Question

Compute the image location and magnification of an object 40 cm from a doublet thin lens
combination having focal lengths f1 = +20 cm and f2 = −60 cm, given that the second
lens is positioned 12 cm behind the front one. If the object is two centimeters high determine
the size and the type of the image. Make a sketch of appropriate rays that support your
findings.

Answer 1

given :

u1 = object distance for first lens = -40 cm

f1 = focal length of first lens = +20 cm

v1 = image distance for first lens

u2 = object distance for second lens

v2 = image distance for second lens

f2 = focal length of second lens = - 60 cm

M = total magnification

ho = height of object = + 2cm

hi = height of image = ?

Using lens equation for first lens :

[u = object distance, v = image distance]\

=> v1 = + 40 cm.

This image will act as object for second lens.

u2 = 40 - 12 = 28 cm [ behind secons lens]

we will use, u2 = + 28 cm

Now

Using lens equation for second lens :

[u = object distance, v = image distance]\

=> v2 = + 52.5 cm.

Therefore, final image formed is REAL.

Total magnification = M = magnification by first lens x magnition by second lens = .

The -ve sign signifies that image formed is INVERTED.

Also, M = height of image / height of object

=> Height of image = M x height of object = -1.875 x +2 = - 3.75 cm [size of image] [answer]

see the ray diagram :