Question

If an object is kept at a distance of 15 cm in front of a diverging mirror of focal length 10 cm . What is the image distance and magnification?​

Answer 1

Given:

An object is kept at a distance of 15 cm in front of a diverging mirror of focal length 10 cm .

To find:

• Image distance
• Magnification

Calculation:

Applying Mirror Formula:

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

$\rm \implies \dfrac{1}{( - 10)} = \dfrac{1}{( - 15)} + \dfrac{1}{v}$

$\rm \implies - \dfrac{1}{10} = - \dfrac{1}{15} + \dfrac{1}{v}$

$\rm \implies \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{10}$

$\rm \implies \dfrac{1}{v} = \dfrac{10 - 15}{150}$

$\rm \implies \dfrac{1}{v} = - \dfrac{1}{30}$

$\rm \implies v = - 30 \: cm$

So, image distance is 30cm in front of mirror.

$\rm \: mag. = - \dfrac{v}{u}$

$\rm \implies\: mag. = - \dfrac{ - 30}{ - 10}$

$\rm \implies\: mag. = - 3$

So, magnification is -3.