Question

find the focul length of a concave mirror which forms an image at a distance of 30 cm when the object of height 6cm is placed at a distance of 20 cm from the mirror. Find the magnification and nature of image.​

Answer 1

$\bold{GIVEN}$

▶Image distance v= - 30cm

▶Object height = 6 cm

▶Object distance u= - 20cm

By mirror's formula we know,

$\bigstar \large \boxed{\mathsf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}$

On applying values,

$\mathsf{\dfrac{1}{f} = \dfrac{1}{-30} + \dfrac{1}{-20}}$

$\mathsf{\dfrac{1}{f} = \dfrac{-2 -3}{60} }$

$\mathsf{\dfrac{1}{f} = \dfrac{-5}{60} }$

$\mathsf{\dfrac{1}{f} = \dfrac{1}{-12}}$

So,

$\large \mathcal{FOCAL \: LENGTH = -12 cm}$

WE KNOW,

$\bigstar \large \boxed{\mathsf{Magnification= \dfrac{v}{u} = \dfrac{Image \: height \:I}{Object \: height \:O}}}$

On placing values,

$\mathsf{Magnification=\dfrac{30}{20} = \dfrac{Image \: height \:I}{6}}$

$\mathsf{\dfrac{30}{20} = \dfrac{Image \: height \:I}{6}}$

$\mathsf{\dfrac{3 \times 6}{2} = Image \: height \:I}$

$\large{\mathcal{9 cm= Image \: height \:I}}$

$\large \underline{\mathcal{NATURE \: OF \: IMAGE}}$

▶Since, the image size is greater than 1 so the image is magnified.

▶Image is formed on same side as of object

▶Image formed will be erect and real.