Question

a 2 centimetre tall object placed at 15 CM in front of a convex mirror of focal length 10 cm determine position size and nature of image​

Answer 1

HEY MATE YOUR ANSWER IS HERE.....

★ GIVEN ★

HEIGHT OF OBJECT ( ho ) = 2 cm

OBJECT DISTANCE ( u ) = - 15 cm

MIRROR USED = CONVEX

FOCAL LENGTH ( f ) = + 10 cm

★ SIGN CONVENTION ★

HO = (+)

U = (-)

F = (+)

V = (+)

★ SOLUTION ★

BY USING MIRROR FORMULA ...

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

HENCE ....

$\frac{1}{10} = \frac{1}{v} - \frac{1}{15}$

$\frac{1}{10} + \frac{1}{15} = \frac{1}{v}$

$\frac{15 + 10}{150} = \frac{1}{v}$

$\frac{25}{150} = \frac{1}{v}$

$v \: = 6 \: cm \:$

NOW BY MAGNIFICATION...

$\frac{hi}{ho} = \frac{ - v}{u}$

$\frac{hi}{2} = - \frac{6}{</strong><strong>-</strong><strong>15}$

hence

$hi \: = \frac{12}{15} \: \: cm$

now BY MAGNIFICATION....

$</strong><strong>magnificat</strong><strong>ion</strong><strong> = \frac{ - v}{u}$

IMAGE IS VIRTUAL , SMALLER , AND ERECT....

THANKS FOR YOUR QUESTION HOPE THIS HELPS...

★ KEEP SMILING ☺️☺️ ★

Answer 2

GIVEN

height of object, H = 2 cm

let, height of image be h

position of object, u = -15 cm

let, position of image be v

focal length of convex mirror, f = 10 cm

SOLUTION

By mirror formula

we know

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \\ so \\ \\ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ (putting \: values) \\ \\ \frac{1}{v} = \frac{1}{10} - \frac{1}{( - 15)} \\ \\ \frac{1}{v} = \frac{3 + 2}{30} \\ \\ \frac{1}{v} = \frac{1}{6} \\ \\ v = 6 \: cm$

now,

as we know

magnification of mirror

m= -v/u = h / H

so,

-(6) / -15 = h / 2

h = (12 / 15) cm

Hence,

the position of image will be 6 cm behind the mirror,

size will be 12/15 cm and nature of image will be virtual and erect.