Question

An object 5cm high is placed at a distance of 10 cm from a convex mirror of radius of curvature 30 cm. Find the nature, position, anf size of the image​

Answer 1

Solution:-

$\underline{\rm \: given}$

$\rm \: \to \: h_0 = 5cm$

$\rm \to \: R = 30cm$

We know that

$\to \: \rm \: f = \dfrac{R}{2} = \dfrac{30}{2} = 15cm$

$\to\rm \: object \: distance \: (u) = 10cm \:$

$\red{\underline {\rm using \: sign \: convention}}$

1) All distance measured from pole

2) Distance measured in the direction of incidents ray are taken as positive

So U is the opposite to Incident ray

$\rm \to\: object \: distance \: (u) = - 10cm \:$

as we know that focal length is always positive in convex lens so

$\rm \: focal \: length \: (f) = + 15cm$

$\red{ \underline{ \rm \: formula}}$

$\implies \boxed{ \rm \: \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }$

$\rm \to \: \dfrac{1}{15} = \dfrac{1}{v} - \dfrac{1}{10}$

$\rm \: \to \: \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{10}$

$\to\rm \: \dfrac{1}{v} = \dfrac{2 + 3}{30}$

$\rm \: \to \: \dfrac{1}{v} = \dfrac{5}{30}$

$\boxed{ \to \rm \: v = + 6cm}$

$\rm \: magnification \: (m \: ) = \dfrac{ - v}{u}$

$\rm \to \: m \: = \dfrac{ - 6}{ -10 } = \dfrac{2}{5}$

$\rm \: magnification \: (m \: ) = \dfrac{ h_i}{h_o}$

$\rm \: \to \: \dfrac{2}{5} = \dfrac{h_i}{5}$

$\rm \: \to \: h_i = 2cm$

$\rm i)position \: of \: the \: image = behind \: the \: mirror$

$\rm \: ii)nature \: of \: image \: = \: virtual \: and \: erect$

$\rm \: iii)size \: of \: image \: = \: smaller \: than \: object$