Question

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

Answer 1

Given :

object height, h = 5cm.

Object distance, u = -10cm.

radius of curvature, R = 30cm.

To find :

the nature, position and size of the image

Solution :

we know that,

focal length = R/2 = 30/2 = 15cm.

Now, using Mirror formula .i.e.,

A formula which gives the relationship between image distance, object distance and focal length of the spherical mirror is known as the mirror formula. i.e.,

${\leadsto{\bf{ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}}$

${\leadsto{\bf{ \dfrac{1}{v} + \dfrac{1}{ - 10} = \dfrac{1}{15} }}}$

${\leadsto{\bf{ \dfrac{1}{v} - \dfrac{1}{10} = \dfrac{1}{15} }}}$

${\leadsto{\bf{ \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{10} }}}$

${\leadsto{\bf{ \dfrac{1}{v} = \dfrac{2 + 3}{30} }}}$

${\leadsto{\bf{ \dfrac{1}{v} = \dfrac{5}{30} }}}$

${\leadsto{\bf{ \dfrac{1}{v} = \dfrac{1}{6} }}}$

${\leadsto{\bf{v = 6cm }}}$

thus, the position of image is 6 cm behind the convex mirror.

To find the size of image, we will calculate the magnification first.

${\leadsto{\bf{magnification(m) = - \dfrac{v}{u} }}}$

${\leadsto{\bf{m = - \dfrac{6}{ - 10} }}}$

${\leadsto{\bf{m = \dfrac{6}{ 10} }}}$

${\leadsto{\bf{m = 0.6}}}$

we also have another formula for magnification, which is :

${\leadsto{\bf{m = \dfrac{h'}{ h} }}}$

${\leadsto{\bf{0.6 = \dfrac{h'}{ 5} }}}$

${\leadsto{\bf{h' = 0.6 \times 5}}}$

${\leadsto{\bf{h' = 3 \: cm}}}$

thus, the size of image is 3 cm.

Nature of image :

• the image will be virtual and erect.
• the image is diminished