Question

An object of size 7.5 cm is placed in front of a convex mirror of radius of curvature 25 cm at a distance of 40 cm.the size o he image should be...........................

Answer 1

size of object, h = 7.5cm
radius of curvature of convex mirror, R = +25cm
focal length, f = R/2 = 25/2 cm
object distance, u = -40cm
image distance = v

we know that
$\frac{1}{v} + \frac{1}{u}= \frac{1}{f} \\ \\ \Rightarrow \frac{1}{v}= \frac{1}{f} - \frac{1}{u}\\ \\ \Rightarrow \frac{1}{v}= \frac{1}{25/2} - \frac{1}{40}\\ \\ \Rightarrow \frac{1}{v}= \frac{2}{25} - \frac{1}{40}\\ \\ \Rightarrow \frac{1}{v}= \frac{16-5}{200}\\ \\ \Rightarrow \frac{1}{v}= \frac{11}{200}\\ \\ \Rightarrow v= \frac{200}{11}cm$

magnification = $\frac{h'}{h} =- \frac{v}{u}$
h' = height of image

$\frac{h'}{h} =- \frac{v}{u}\\ \\ h' = -h \times \frac{v}{u}=-7.5 \times \frac{200}{11 \times (-40)}= \frac{75}{22}\ cm=3.41cm$

Size of image is 3.41cm.

Answer 2

size of object, h = 7.5cm

radius of curvature of convex mirror, R = +25cm

focal length, f = R/2 = 25/2 cm

object distance, u = -40cm

image distance = v

we know that

magnification = 

h' = height of image

Size of image is 3.41cm.