Question

Find the size nauter and position of image formed by concave mirror when an object of a size 1cm is placed at a distance of 15cm given focal length of mirror is 10cm

Answer 1

Given :-

• Object distance (u) = -15 cm

• Focal length (f) = -10 cm

• Height of object (h1) = 1 cm

To find :-

• Size,nature and position of image formed by concave mirror.

Solution :-

$\blue{\bigstar}\:\:\underbrace{\bf\gray{{By\:using\:mirror\:formula}}}$

$\purple{\bigstar}\:\:{\underline{\boxed{\bf\red{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}$

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

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

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

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

$\rm\longrightarrow\:\dfrac{1}{v}=-\dfrac{1}{30}$

$\rm\longrightarrow\:v=-30$

❍  Hence,the position of the image is 30 cm .Minus sign shows the left side of mirror

❍  Since the image is formed in front of the concave mirror it's nature will be real and inverted.

Now,

${\underline{\boxed{\bf\pink{Magnification\:(m)=-\dfrac{v}{u}}}}}$

$\rm\longrightarrow\:m=-\dfrac{(-30)}{(-15)}$

$\rm\longrightarrow\:m=-\cancel\dfrac{-30}{-15}$

$\rm\longrightarrow\:m=-2$

Now,

${\underline{\boxed{\bf\purple{m=\dfrac{h_2}{h_1}}}}}$

$\rm\longrightarrow\:-2=\dfrac{h_2}{1}$

$\rm\longrightarrow\:h_2=-2\times\:1$

$\rm\longrightarrow\:h_2=-2\:cm$

Hence,the size of image is 2 cm long.(Minus sign show that the image is formed below the principal axis) and it is a real and inverted image.