Question

A particle goes in a circle of radius 2.0 cm. A concave mirror of focal length 20 cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. Calculate the radius of the circle formed by the image.

Answer 1

Radius of the image of the circle is 4 cm

Explanation:

u = -30 cm

f = -20 cm

By mirror equation: 1/v + 1/u = 1/f

1/v - 1/30 = -1/20

1/v = 1/30 - 1/20

 1/v = 2/60 - 3/60

1/v = - 1/60

Therefore v = -60 cm

Image of the circle is formed 60 cm in front of the mirror, same side as the object.

magnification m = - v/ u = - (- 60 / -30)  

R(image) / R(object) = m

So R(image) / 2 = - 60/30

R (image) = 60 * 2 / 30 = 4cm

Radius of the image of the circle is 4 cm

Answer 2

The radius of the circle formed by the image is 4 cm.

Explanation:

Given data in the question  

A particle goes in a radius of 2.0 cm

           u = -30cm

           f = -20cm

Here, f is a concave, focal length mirror .

         u is the distance between the mirror pole and circle centre .

We know that equation of mirror,

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

         $\frac{1}{v}+\frac{-1}{30}=-\frac{1}{20}$

         $\frac{1}{v}=\frac{1}{30}-\frac{1}{20}$

         $\frac{1}{v}=\frac{1-2}{60}$

         $\frac{1}{v}=-\frac{1}{60}$

Hence, $v$ =  - 60 cm

Circle image is shaped 60 cm away in front of the mirror :

          $m=-\frac{v}{u}=\frac{R_{\text {image}}}{R_{\text {object}}}$

              $=-\frac{(-60)}{-30}=\frac{r_{i m a g e}}{2}$    

    $R_{\text {image}}=4 \mathrm{cm}$

Therefore, the circle image radius is 4 cm.