Question

10. An object of size 2cm is placed at distance 20 cm from a concave
mirror of focal length 30 cm. At what distance a screen must
placed to get a sharp and clear image and find the size of the
image.​

Answer 1

Explanation:

M = F/( F - U)

sign convention,

M = - 30 / ( -30 +20)

= - 30 / ( -10)

= 3

M = 3

then,

M =size of image / size of object =3

3 = image /2

size of image = 6cm

M = - V/U

6 = V/ 20

V = 120 cm.

Answer 2

$\bf \: \underline{Given} :$ ↴

$\sf \: \implies \: h_o = 2 \: cm$

$\sf \:\implies \: u = - 20 \: cm$

$\sf \:\implies \: f = - 30 \: cm$

$\bf \: \underline{To \: find } :$ ↴

$\sf \:\implies \: image \: distance (v)$

$\sf \:\implies \: height \: of \: image \: ( v_i)$

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

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

$\: \implies \sf\dfrac{1}{ - 30} = \: \dfrac{1}{v} + \dfrac{1}{ - 20}$

$\: \implies \sf\dfrac{1}{ - 30} + \dfrac{1}{20} = \: \dfrac{1}{v}$

$\: \implies \sf\dfrac{ - 2 + 3}{ 60} = \: \dfrac{1}{v}$

$\implies \sf\dfrac{1}{ 60} = \: \dfrac{1}{v}$

$\red{ \implies \sf v = 60 }$

So, the screen must be placed at the distance of 60 cm from the mirror.

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

$\implies \sf \: m = \dfrac{ - 60}{ - 20}$

$\implies \sf \: m = \dfrac{ \!\!\!- \!\!\!\not6\!\!\!\not0}{ - \!\!\!\not2\!\!\!\not0}$

$\red{\implies \sf \: m = 3}$

So, the magnification (m) is 3

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

$\implies \sf \: 3 = \dfrac{h_i}{2}$

$\implies \sf h_i = 3 \times 2$

$\red{\implies \sf h_i =6 \: cm}$

Hence, the height of the image is 6 cm

________________________________

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

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

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

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

$\sf \: R = Radius \: of \: curvature$

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