Question

If the focal length of a concave lens is 15cm, when the object is placed at a distance of 25cm from the lens, then determine the height of the image formed if the height of the object is 4cm​

Answer 1

Answer:

let distance of object be u,image be v,height of object be Ho and height of object be He.

u=-25cm

f=15cm

then

1/f=1/v-1/u

1/v=1/f+1/u

=1/15-1/25

=(5-3)/75

=2/75

then

v=75/2

=37.5cm

He/Ho=v/u

He=v/u*Ho

=37.5/-25*4

=-6

height of image is 6cm inverted.

I hope it will help you.

Thanks

Answer 2

Answer:

Explanation:

Given,

Focal length, f = 15 cm

Object distance, u = - 25 cm

Object height, h(o) = 4 cm

To Find,

Height of the image, h(i)

Formula to be used,

Lens formula,

1/v - 1/u = 1/f

Magnification formula,

m = v/u

Solution,

Putting all the values, we get

1/v - 1/u = 1/f

⇒ 1/v - 1/(- 25) = 1/15

⇒ 1/v + 1/25 = 1/15

⇒ 1/v = 1/15 - 1/25

⇒ 1/v = (5 - 3)/75

⇒ 1/v = 2/75

⇒ v = 75/2

⇒ v = 37.5 cm

Here, the image distance is 37.5 cm.

Now, We know that,

m = v/u,

⇒ m = 37.5/- 25

⇒ m =- 1.5

Then,

m = h(i)/h(o)

⇒ - 1.5 = h(i)/4

⇒ h(i) = 4 × (- 1.5)

⇒ h(i) = - 6 cm

Hence, the image height is - 6 cm.