Question

An object of height 3 cm is placed at a distance of 10cm from a convex mirror of focal length 15 cm . Find the position, size and nature of the image.

Answer 1

$\bf \: \underline{Given} :$

$\sf \implies \: h_o = 3 \: cm$

$\sf \implies \: u = - 10 \: cm$

$\sf \implies \: f = 15 \: cm$

$\bf \: \underline{To \: find } :$

$\sf\implies Position, size \: and \: nature \: of \: the \: image.$

$\bf \underline{ \underline{Solution }}$

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

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{f} - \dfrac{1}{u}$

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

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

$\implies \sf \dfrac{1}{v} =\: \dfrac{2}{30} + \dfrac{3}{30}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{2 + 3}{30}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{ 5}{30}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{6}$

$\implies \sf v = 6$

Image distance = 6 cm

$\bf Now,$

$\sf Finding \: the \: height \: of \: the \: image,$

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

$\implies \sf \: m = \dfrac{ - 16}{ - 10}$

$\implies \sf \: m = \dfrac{3}{5}$

$\implies \sf \: m =0.6$

So, the magnification (m) is 0.6

$\implies \boxed{ \pink{\sf \: m = \dfrac{h_i}{h_o} }}$

$\implies \sf \: 0.6 = \dfrac{h_i}{3}$

$\implies \sf h_i = 0.6\times 3$

$\red{\implies \sf h_i =1.8 \: cm}$

Hence, the height of the image is 1.8 cm and the image formed is erect and virtual.

________________________________

$\sf \: h_o = Height \: of \: object$

$\sf \: h_i = Height \: of \: image$

$\sf \: u = Object \: distance \: from \: mirror$

$\sf \: v = image \: distance \: from \: mirror$

$\sf \: f = focal \: length \: of \: mirror$

Answer 2

Explanation:

$\bf \: \underline{Given} :$

$\sf \implies \: h_o = 3 \: cm$

$\sf \implies \: u = - 10 \: cm$

$\sf \implies \: f = 15 \: cm$

$\bf \: \underline{To \: find } :$

$\sf\implies Position, size \: and \: nature \: of \: the \: image.$

$\bf \underline{ \underline{Solution }}$

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

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{f} - \dfrac{1}{u}$

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

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

$\implies \sf \dfrac{1}{v} =\: \dfrac{2}{30} + \dfrac{3}{30}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{2 + 3}{30}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{ 5}{30}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{6}$

$\implies \sf v = 6$

Image distance = 6 cm

$\bf Now,$

$\sf Finding \: the \: height \: of \: the \: image,$

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

$\implies \sf \: m = \dfrac{ - 16}{ - 10}$

$\implies \sf \: m = \dfrac{3}{5}$

$\implies \sf \: m =0.6$

So, the magnification (m) is 0.6

$\implies \boxed{ \pink{\sf \: m = \dfrac{h_i}{h_o} }}$

$\implies \sf \: 0.6 = \dfrac{h_i}{3}$

$\implies \sf h_i = 0.6\times 3$

$\red{\implies \sf h_i =1.8 \: cm}$

Hence, the height of the image is 1.8 cm and the image formed is erect and virtual.

________________________________

$\sf \: h_o = Height \: of \: object$

$\sf \: h_i = Height \: of \: image$

$\sf \: u = Object \: distance \: from \: mirror$

$\sf \: v = image \: distance \: from \: mirror$

$\sf \: f = focal \: length \: of \: mirror$