Question

the above lens has a focal length of 10cm the object of height 2mm is placed at a distance of 5cm from the pole.find the heights of image​

Answer 1

Explanation:

f=10cm, h=2mm = 1/5cm and u=-5cm

Now, we know that,

1/f= 1/v + 1/u

1/v= 1/f - (-1/u)

1/v= 1/10 + 1/5

1/v = 1+2/10

1/v = 3/10

v= 10/3cm

Now, magnification

(m)= h'/h = -v/u

h' = -10/3

__. _____

1/5. -5

Therefore, h'= 1/15cm

Answer 2

Given: The lens has a focal length of 10 cm. The object of height 2 mm is placed at distance of 5 cm from the pole.

To find: Height of the image

Explanation: Focal length of the lens= 10 cm

Height of object= 2 mm = 0.2 cm

Object distance= -5 cm(object distance is always negative)

Using lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

$\frac{1}{v} = \frac{1}{10} + \frac{1}{ - 5}$

$\frac{1}{v} = \frac{1 - 2}{10}$

$\frac{1}{v} = - \frac{1}{10}$

v= -10 cm

Magnification= v/u

= -10/-5

= 2

Also, magnification= height of image/ height of object

=> 2 = height of image/0.2 cm

=> height of image= 0.4 cm

Therefore, the height of the image is 0.4 cm or 4 mm.