Question

8. When an object of height 1 cm is kept at a distance 4 cm from a concave mirror, its erect image of height 1.5 cm is
formed at a distance 6 cm behind the mirror. Find the focal length of mirror, by drawing

Answer 1

Answer:

answer in explaination

Answer 2

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given:- }}}$

$\sf\implies Height \ of \ object \ is \ 1cm. \\\sf\implies Height \ of \ image\ is \ 1.5cm ( erect ) . \\\sf\implies Image \ is\ formed \ 6cm \ behind\ Mirror. \\\sf\implies Object \ distance \ is \ 4cm.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\sf\implies The \ focal \ lenght \ of \ the \ mirror.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

We need to find the focal length of the concave mirror. So we know the relation between magnification , focal length & image Distance / Object distance as ,

$\underline{\boxed{\purple{\bf \pink{\leadsto} \: m = \dfrac{f-v}{f} = \dfrac{ f}{f-u} }}}$

So , we can either use image distance or object distance to find the focal length . Here I will be using the image distance. Since the image is formed behind the mirror , hence it will be positive.

$\begin{lgathered}\sf:\implies \pink{ m = \dfrac{ f - v }{f}}\\\\\sf:\implies \dfrac{1.5cm}{1cm} = \dfrac{ f - 6cm }{f }\\\\\sf:\implies 1.5 = \dfrac{f-6cm}{f}\\\\\sf:\implies 1.5f = f - 6cm \\\\\sf:\implies 1.5f - f = -6cm \\\\\sf:\implies 0.5 f = -6cm \\\\\sf:\implies f = \dfrac{ -6cm \times 10}{5} \\\\\sf:\implies \boxed{\pink{\mathfrak{ Focal \ lenght = -12 cm }}}\end{lgathered}$

$\underline{\blue{\sf \therefore Hence \ the \ focal \ length \ is \ \textsf{\textbf{-12 \ cm }}. }}$