Question

An object of size 4 cm is placed 60 cm away from a converging mirror of radius of curvature 1 m calculate the position nature and size of the image.​

Answer 1

Explanation:

hope this will help you.

Answer 2

Given:

size of object h = 4 cm

radius of curvature R = 1 m

distance between object u = - 60 cm

To find:

1) position =?

2) nature =?

3) size of the image =?

Step-to-step-explanation:

• At first, the focal length is calculated which is half of the radius of curvature.

         f = 0.5 × R

           = 0.5 × 100

         f = 50 cm

• By the mirror's formula,

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

        $\displaystyle \frac{1}{v} +\frac{1}{- 60} =\frac{1}{50}$

       v = 27.2727 cm

• The magnification is the ratio of image height & object height as well as the ratio of image distance & object distance.

       $\displaystyle m=\frac{-v}{u} =\frac{h_i}{h_o}$

• Put given values in the above equation to find the height of the image.

      $\displaystyle \frac{-27.27}{-60} = \frac{h_i}{4}$

       $h_i$ = 1.818 cm

1)

Position of Image formed: Behind the mirror.

2)

Nature of Image formed: Virtual and Erect.

3)

Size of the Image formed: Diminished.