Question

am object of 3cm high is placed at a distance of 8cm from which producesa virtual image of 4.5 high
1)what is the focal length of the mirror.2)what is the position of the image​

Answer 1

object height, h1 = 3cm

image height, h2= 4.5cm

image distance,u= 8cm

object distance, v = ?

m= h2/h1= -v/u

4.5/3= -v/8

-v = 12

v = -12cm

1) focal length=

1/f = 1/v + 1/u

1/f = 1/-12 + 1/8

1/f= -2+3/24

1/f = 1/24

f= 24cm

2) as the focal length is +ve it is a convex mirror.
so the position of the image is at infinity
idk the second one correctly

Answer 2

$\huge{\mathfrak{Question:-}}$

An object of 3cm high is placed at a distance of 8cm from which producesa virtual image of 4.5 high. Find:

→What is the focal length of the mirror.

→What is the position of the image​

$\huge{\mathfrak{Solution:-}}$

$\large{\textsc{Given:-}$

$\bold{ho = 3\,cm}$

$\bold{u = -8\,cm}$

$\bold{hi = 4.5\,cm}$

Clearly, the object is placed between f and p.

Now, magnification;

$\bold{m = \frac{h_{i} }{h_{o}} = \frac{v}{u}}\\ \\ \bold{= \frac{4.5}{3} = \frac{-v}{-8}}\\\\\bold{\implies{v = 12\,cm}}$

Thus,

By mirror formula we find the focal length

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

$\bold{\frac{1}{f} = \frac{1}{12} +\frac{1}{-8}}\\ \\\bold{\frac{1}{f} = \frac{1}{12} -\frac{1}{8}}$

Take LCM we get,

$\bold{= \frac{2-3}{24}}\\ \\ \bold{=\frac{-1}{24}}\\ \\ \bold{\boxed{\boxed{\bold{f = -24\,cm}}}}$

Hence, focal length of mirror = -24cm

and postion of the image is between f and p.

$\large{\textsc{Extra\; Info:-}$

$\bold{1). Mirror\; Formula=\frac{1}{f} = \frac{1}{v}+\frac{1}{u}}$

$\bold{2). Lens\; Formula = \frac{1}{f} =\frac{1}{v}- \frac{1}{u}}$

$\bold{3). Magnification = \frac{h_{i} }{h_{o} } =\frac{v}{u}}$