Question

2. An object is placed at 10 cm in front of a concave mirror of focal length 7.5 cm.Find the position,nature and magnification of the image.
(-30cm,real & inverted,-3)

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Object is placed at (u) = -10cm
• Focal Lenght (f) = -7.5cm

Object Distance and focal Lenght are negative because they are at left side of mirror.

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Position of image (v)
• Nature of image
• Magnification of image

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

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

$\implies\underline{\boxed{\sf Magnification\:(m)=\dfrac{-v}{u}}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

$\implies{\sf \dfrac{1}{-7.5}=\dfrac{1}{v}+\left(-\dfrac{1}{10}\right) }$

$\implies{\sf \dfrac{1}{-7.5}=\dfrac{1}{v}-\dfrac{1}{10}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{10+(-7.5)}{-7.5 \times 10} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{2.5}{-75}}$

$\implies{\sf v=\dfrac{-75}{2.5} }$

$\implies{\bf v=-30\:cm}$

$\bullet\large{\bf Position \: and \: nature \:of image -}$

• Image is formed between infinite and radius of curvature
• Image is large
• Image formed is real and erect / inverted

$\bullet\large{\bf Magnification \: of \: image -}$

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

$\implies{\sf m=\dfrac{-30}{10} }$

$\implies{\bf m=-3}$

Magnification of image is -3