Question

An object is placed at a distance of 30 cm from a concave mirror of focal length 15 cm.

Answer 1

Given:

$\sf{u=-30cm}$

$\sf{f=-15cm}$

$\sf{Mirror=Concave}$

Solution:

$\sf{{\dfrac{1}{f}}={\dfrac{1}{v}}+{\dfrac{1}{u}}}$

$\sf{-{\dfrac{1}{15}}={\dfrac{1}{v}}-{\dfrac{1}{30}}}$

$\sf{{\dfrac{1}{v}}={\dfrac{1}{30}}-{\dfrac{1}{15}}}$

$\sf{{\dfrac{1}{v}}={\dfrac{1-2}{30}}}$

$\sf{{\dfrac{1}{v}}=-{\dfrac{1}{30}}}$

$\sf{v=-30cm}$

$\bold{Magnification:}$

$\sf{m=}{\dfrac{-v}{u}}$

$\sf{m=-}{\dfrac{-30}{-30}}$

$\sf{m=-1}$

Thus, the image will form at the same place where the object is placed.

Also, the image will be real and inverted and of the same size as object.