Question

An object 5 cm in length is placed 25 cm away from
converging lens of focal length 10 cm. Find the position
size and nature of the image formed.
​

Answer 1

Explanation:

given,u=-25cm,f=10cm,h=5cm,v=?,h'=?

1/v-1/u=1/f

1/v=1/f+1/u

=1/10+1/-25

=1/10-1/25

=5-3/50

=2/50

v=50/3

=16.67cm

again,h'/h=v/u

h'=v×h/u

=16.67×5/-25

=-3.3cm

position of the image 16.67cm behind the converging lens

size of the image 3.3cm, diminished

nature of the image real and inverted

Answer 2

Hello Dear Soul ❤

Your Answer :

Height of the object, h' = 5 cm

Distance of object from conversing lens, u = 25 cm

Focal length of conversing lens, f = 10 cm

Using lens formula,

1/v - 1/u = 1/f

1/v = 1/f + 1/u

1/v = 1/10 - 1/25 = 15/250

v = 250/15 = 16.66 cm

Also for conversing lens,

h''/h' = v/u { h'' = image height }

h'' = (v/u)× h' = (16.66 × 5) / (-25) = -10/3 = -3.3 cm

Thus, The image is inverted and formed at a distance of 16.7 cm behind the lens and measured 3.3 cm.

Note : Ray diagram of this problem is given in attached file.

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