Question

The radius of curvature of a concave mirror is 30cm an object of length 5cm is kept at a distance of 10cm in front of the mirror find the position length of nature of image

Answer 1

Given :

• Radius of Curvature (R) = -30cm
• Object Lenght (ho) = 5cm
• Distance of object from mirror (u) = -10cm

Sign convention -

Radius of Curvature and object are at left side of concave mirror therefore they are taken as negative

To Find :

• Position and length of image
• Nature of image

Formula Used :

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

Solution :

$\implies{\sf R = 2f }$

$\implies{\sf f = \dfrac{-30}{2}}$

$\implies{\bf f = -15 \; cm }$

__________________________________

$\implies{\sf \dfrac{1}{-15}=\dfrac{1}{v}+\left(\dfrac{1}{-10}\right) }$

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

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

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

$\implies{\sf v =\dfrac{-150}{-5} }$

$\implies{\bf v = +30\; cm}$

Nature of Image -

• Image formed behind the mirror
• Image size is large
• Virtual , erect

__________________________________

➞ Lenght of image -

$\implies{\sf m =\dfrac{h_i}{h_o}=\dfrac{-v}{u} }$

$\implies{\sf \dfrac{h_i}{5}=\dfrac{30}{10} }$

$\implies{\sf h_i = 3 \times 5}$

$\implies{\sf h_i = 15 \: cm }$

Answer :

Distance of Image from mirror = +30cm

Height of image (hi) = 15cm