Two converging lenses, with focal length f1=20cm and f2=30cm, are placed 90cm apart. An object of the height 5 cm is placed 50 cm in front of the first lens with f1=20cm. Determine the position, of the final image formed by the combination of the two lenses. Draw the ray diagram.

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The image of the object formed by the first lens alone has position and height,

`v=(uf_1)/(u-f_1)` ,where v is distance of the image from lense , u is distance of the object from lense ,`f_1` is focal lenth of first lense.

Thus `f_1=20 cm ,u=50 cm ,h= 5 ` cm hieght of the object.

`v=(50xx20)/(50-20)=1000/30=33.33 ` cm

Between first and second lense ,56.67 cm away from second lense.

height of the image

`h_1=-(v/u)h`

`=-(33.33/50)xx5 cm`

`=-3.33 cm`

(inverted image which is object for second lense)

(`u> 2f_1 ,` image real inverted and reduced ) (i)

Now u = 56.67 cm ( for second lense)

`f_2=30cm` (given)

`v` (second lense)=`(uf_2)/(u-f_2)`

`=(56.67xx30)/(56.67-30) cm`

`=63.75 cm`

Thus final image behind the second lense ,63.75 cm away ,real ,inverted and enlarge .

(`f_2 <u<2f_2` , image is real ,inverted and enlarge) (ii)

hieght of the second image

`h_2=-(v/u)(h_1)`

`=-(63.75/56.67)(-3.33) cm`

`=3.75 cm`

real and errect.

Thus final image of the object is real ,errect and demnished.

Position of the object w.r.t. 1st lens is beyond 2F.

Position of the image of the object from 1st lens,

v = 50*20/(50-20) = 33.33 cm. It will be real and inverted.

Position of this image which will be the object w.r.t. 2nd lens, is at a distance of 56.67 cm from it. Therefore, it is between F and 2F.

Position of its image w.r.t. 2nd lens,

v = 56.67*30/(56.67-30) = 63.75 cm. It will be real and inverted.

So, after two stage inversion, the last image will be erect, at a distance of 63.75 cm, from the second lens and at its backside.

The ray diagram is attached.

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