Question

A convex mirror produces an image which is 1/4 the size of the object.if the radius of curvature of mirror is 30cm,the distance between the object and it's image is?

Answer 1

Answer:

-75cm

Explanation:

magnification (m) = 1/4

m = -v/u

1/4 = -v/u

radius of curvature = 30cm

therefore,  focal length (f) = 15cm

u = -4u      ----------- Equation (1)$\frac{1}{f} = \frac{1}{v} +\frac{1}{u} \\\\\frac{1}{15} = \frac{1}{v}+\frac{1}{-4v} \:\:\:\:\:------ from \:equation (1)\\\\\frac{1}{15} = \frac{4-1}{4v} \\\\\frac{1}{15}= \frac{3}{4v}\\\\v = \frac{3 \times15}{4} \\\\v = \frac{75}{4} \\\\from\: equation (1)\\u=-4v\\u=- 4\times\frac{75}{4} \\u = -75 cm$

HOPE YOU UNDERSTOOD THE SOLUTION