Question

10. The image of an object placed at a distance of
30 cm on the principal axis of a concave mirror
from its pole, is formed on the object itself. Find
(a) the focal length and (b) linear magnification
of mirror​

Answer 1

Explanation:

HOPE IT HELPS!!

GOOD LUCK

Answer 2

★ Solution:-

• image distance = - 30 cm
• object distance = - 30 cm

The image and the object distance are the same  . Both the distances are negative because distances are measured from the pole of the mirror

First let's find the focal length of the mirror in order to do that simply use the mirror formula ,

$\longrightarrow\mathtt{\dfrac{1}{f} = \dfrac{1}{v}+ \dfrac{1}{u}}$

$\longrightarrow\mathtt{\dfrac{1}{f} = \dfrac{1}{-30}+ \dfrac{1}{-30}}$

$\longrightarrow\mathtt{\dfrac{1}{f} = \dfrac{-1}{30}-\dfrac{1}{30}}$

$\longrightarrow\mathtt{\dfrac{1}{f} = \dfrac{-1-1}{30}}$

$\longrightarrow\mathtt{\dfrac{1}{f} = \dfrac{-2}{30}}$

$\longrightarrow \boxed{\mathtt{f = - 15 cm }}$

(as the focal length of concave mirror is always negative )

The focal length of the concave mirror is 15 cm .

Magnification is given by the formula ,

$\longrightarrow\mathtt{m= -\dfrac{v}{u}}$

$\longrightarrow\mathtt{m= -\dfrac{-30}{-30}}$

$\longrightarrow\boxed{\mathtt{m= - 1 }}$

The linear magnification of the mirror is 1