Question

an object 5 cm high is held 25cm away from a converging lens of focal length 10cm find the position size and nature of the image formed also draw the ray diagram
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Answer 1

Answer:

Given, object size, height of object, h(o) = 5cm

Image size , h(i) = ?

Object distance, u = -25 cm

Image distance, v = ?

Focal length = f = 10cm

From lens formula, 1/f = 1/v - 1/u

1/v = 1/f - 1/u

= 1/10 + 1/(-25)

= 3/50

so, v = 50/3

= 16.67cm from the lens

size of image,

h(i) = h(o) × v/u

= 5 × 50/ 3(-25)

= 3.33cm

{diminished inverted image}

The image is real in nature.

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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