Question

A child at an amusement park stands in front of a concave mirror with a focal length of 75 cm. The child holds her cotton candy close to the mirror and observes that its image is magnified by a factor of five. Determine the distance of the cotton candy from the mirror. What is the radius of curvature of the mirror. If child holds the candy at a distance equal to radius of curvature what is the position, nature and size of the image obtained?

Answer 1

Answer:

The radius of curvature is 150 cm.

The child holds the candy at a distance of 150cm.

The position of the image is at the centre of curvature, it's nature is real and inverted, and the size is equal to that of object.

Answer 2

Answer:

Mirror: concave

Focal length: -75 cm

(i) When child holds the candy close to the mirror(image magnified by 5 times)

magnification, m= 5 (Positive because the image formed is virtual mirror)

m= -v/u

$5=\frac{-v}{u}$

$v= - 5u$------------------(1)

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

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

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

$\frac{1}{-75} =\frac{1}{-5u} +\frac{1}{u}$

on simplifying above equation we get,  u= - 60 cm-------------(2)

from 1 and 2

v = - 5u = -5 x (- 60) = 300 cm

i.e. the image will be formed behind the mirror

(ii) Radius of curvature of the mirror, R = 2f = 2 x 75 = 150 cm

(iii)Child holds the candy at distance equal to radius of curvature(R):

object distance, u= Radius of curvature

When object is placed at the centre of curvature image will also be formed at the centre of curvature.

Therefore, image distance, v = Radius of curvature

Nature of image: Real , inverted, same size as that of the object

Explanation: