Question

A 2.0 cm tall object is placed perpendicular to the principal axis of a

concave lens of focal length 10 cm. The distance of the object from the

mirror is 15 cm. Find the nature, position and size of the image formed.

Represent the situation with the help of a ray diagram.

Thnx :)

Answer 1

Answer:

The image is formed 6 cm in front of the lens.

The image is virtual and erect.

The image is smaller than the object. Its size is 0.8 cm.

For the ray diagram, refer to the attachment.

Explanation:

Given:

• Height of the object, ho = 2 cm
• Focal length, f = - 10 cm
• Object distance, u = -15 cm

To find:

• Nature
• Position
• Size of image

Variables needed:

• Image distance, v = ?
• Magnification, m = ?
• Height of the image, hi = ?

Solution:

• By using lens formula, we get

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\\\\\\\implies \dfrac{1}{-10} = \dfrac{1}{v} - \dfrac{1}{-15}\\\\\\\implies - \dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{15}\\\\\\\implies \dfrac{1}{v} = - \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{-15-10}{150} = \dfrac{-25}{150}\\\\\\\implies \dfrac{1}{v} = - \dfrac{1}{6}\\\\\\\implies \boxed{\bold{v = -6 \; \; cm}}$

Therefore, the image is formed 6 cm in front of the lens.

Now,

$m = \dfrac{v}{u} = \dfrac{-6}{-15} = \dfrac{2}{5}$

Since m is +ve,

The image is virtual and erect.

Also,

$m = \dfrac{hi}{ho}\\\\\\\implies \dfrac{2}{5} = \dfrac{hi}{2}\\\\\\\implies hi = \dfrac{4}{5} = \dfrac{8}{10}\\\\\\\implies \boxed{\bold{hi = 0.8 \; \; cm}}$

Therefore, the image is smaller than the object. Its size is 0.8 cm.

For the ray diagram, refer to the attachment.