Question

If a object of 10cm height is placed at a distance of 36cm from a concave mirror of focal lenght 12cm find the position, nature and height of the image.

Answer 1

Given :

➩ In Concave Mirror

• Object Height (ho) =10cm
• Distance of object (u) = -36cm
• Focal Lenght (f) = -12cm

sign convention

Focal Lenght and object are at left side of mirror therefore they are negative (-)

To Find :

• Position of image
• Height of image

Formula Used :

Mirror's Formula

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

$\bullet\underline{\boxed{\sf m=\dfrac{h_i}{h_o}=\dfrac{-v}{u}}}$

hi = Height of image

ho = Height of object

Solution :

$\implies{\sf \dfrac{1}{-12}=\dfrac{1}{v}+\left(\dfrac{1}{-36}\right) }$

$\implies{\sf \dfrac{1}{-12}=\dfrac{1}{v}-\dfrac{1}{36} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{1}{-12}+\dfrac{1}{36} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{36+(-12)}{-12 \times 36}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{24}{-432}}$

$\implies{\sf v =\dfrac{-432}{24}}$

$\implies{\bf v = -18 \: cm}$

Characteristics of image

• Image formed between infinite and centre of curvature
• Real and erect

➞ Height of image :

$\implies{\sf \dfrac{h_i}{10}=\dfrac{-18}{36}}$

$\implies{\sf h_i=\dfrac{10\times - 1}{2} }$

$\implies{\bf h_i=-5\: cm}$

Answer :

Position of image (v) = -18cm

Height of image (hi) = -5 cm