# 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...

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.

llltkl | College Teacher | (Level 3) Valedictorian

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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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pramodpandey | College Teacher | (Level 3) Valedictorian

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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.