Question

An object 0.04m high is placed in front of and ar righ angle to the axis of the concave mirrior which has radius of converture 0.15m. A)if a real image 0.12m high is obtained ,deter the distance of the object From the mirrior.

Answer 1

$\boxed{\huge{\bf{\red{Answer}}}}$

-2.58 m

Explanation:

$Given \ object \ height \ (h_o)=0.04m\\\\radius \ of \ converture \ (r) =0.15m\\\\and \ image \ height \ (h_e)=0.12m\\\\we \ know \ that\\\\focal \ length \ (f) =\dfrac{radius \ of \ converture \ (r) }{2}\\\\focal \ length \ (f) =\dfrac{ 0.15 }{2}m=0.075m\\\\ we \ know \ that\\\\\dfrac{v}{u} =\dfrac{h_e}{h_o}\\\\putting \ values \ here\\\\\dfrac{v}{u} =\dfrac{0.12}{0.04}=v=0.03u\\\\we \ have \ mirror \ formula\\\\\dfrac{1}{f} =\dfrac{1}{u} +\dfrac{1}{v}\\\\have \ get \ f=0.075m \ and \ v=0.03u$

$we \ have \ to \ find \ object \ distance \ (u)\\\\\\putting \ in \ formula\\\\\\-\dfrac{1}{0.075}=\dfrac{1}{0.03u}+ \dfrac{1}{u}\\\\\\-\dfrac{1000}{75}=\dfrac{100}{3u}+ \dfrac{1}{u}\\\\\\-\dfrac{40}{3}=\dfrac{100+3}{3u}\\\\\\-\dfrac{40}3}=\dfrac{103}{3u}\\\\\\ u=-\dfrac{103}{40}\\\\\\u=-2.575m\\\\\\Thus, \ object \ distance \ is \ 2.58m.$