Question

an object 5 m high is placed at a distance of 10 cm from a convex mirror of radius of curvature 30 cm .find the nature ,position and size of the image .

Answer 1

the nature of image is virtual and erect larger than object the position of image is 6 centimetre from the pole the size of image is 2 metre

Answer 2

Answer:

$Given, \: Radius \: of \: curvature \: of \: a \: convex \: mirror \: (R) \: = + 30cm.</p><p>$

$∴ \: \: \: \: \: \: \: \: \: \: \: Focal \: length \: (ƒ) = \frac{R}{2} = \frac{ + 30}{2} = + 15cm.$

$Object \: distance \: ( u) = - 10cm.</p><p> \\ Height \: of \: object \: ( h) = 5cm.$

$We \: have, \: \: \: \: \: \: \: \: \: \: \frac{1}{ ƒ} = \frac{1}{v} + \frac{1}{u}$

$⇒ \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{1}{ ƒ} - \frac{1}{u} \\ \\ \: \: \: \: \: \: \: \: \: = \frac{1}{15} - ( - \frac{1}{ 10} ) \\ \\ \: \: \: \: \: \: \: \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{1}{6}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: v = 6cm.$

$Magnification, \: \: \: \: \: \: m = - \frac{v}{u} = \frac{ {h}^{'} }{h}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: - ( \frac{6}{ - 10} ) = \frac{ {h}^{'} }{5}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{3}{5} = \frac{ {h}^{'} }{5}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: {h}^{'} = 3cm$

$Image \: will \: be \: virtual \: and \: erect.$