Question

a 6 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 25cm. the distance of the object from lens is 40cm. by calculation determine the position and the size of image formed

Answer 1

Answer: The height of image is 10 cm real and inverted and the image is formed at 66.66 cm on the other side of the lens.

Explanation:

Given that,

Height of the image h = 6 cm

Focal length f = 25 cm

The distance of the object u = -40 cm

Using lens's formula

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

$\dfrac{1}{25}=\dfrac{1}{v}-\dfrac{1}{-40}$

$\dfrac{1}{v}= \dfrac{3}{200}$

$v = \dfrac{200}{3}$

$v = 66.66\ cm$

The image is formed at 66.66 cm.

The magnification is

$m =\dfrac{v}{u} = \dfrac{h'}{h}$

$\dfrac{h'}{6}= \dfrac{\dfrac{200}{3}}{-40}$

$h' = -10 cm$

Hence, The height of image is 10 cm real and inverted and the image is formed at 66.66 cm on the other side of the lens.

Answer 2

Answer:

Explanation:

Given :-

Object Height = 6 cm

Focal length = - 25 cm

Object distance = - 40 cm

Solution :-

Using Lens formula,

1/f = 1/v - 1/u

1/v = 1/f + 1/u

⇒ 1/v = 1/25 + 1/- 40

⇒ 1/v = 8 - 5/200

⇒ 1/v = 3/200

⇒ v = 200/3

⇒ v = 66.67 cm

Hence, the position of the image is 66.67 cm.

Magnification, m = h'/h = v/u

h' = v/u × h

⇒ h' = 200/3 × (- 40) × 6

⇒ h' = - 10 cm.

Hence, the size of image formed is  - 10 cm..