Question

An object 2cm in size is placed 30cm infront of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image?What is the nature and size?

Answer 1

$\bold{GIVEN}$


$\mathsf{Object \: distance = - 30 cm}$

$\mathsf{Focal \: length= - 15 cm}$.

$\mathsf{Image \: distance = ? cm}$

By mirrors formula we know,

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

$\mathsf{\dfrac{1}{-15} = \dfrac{1}{v} + \dfrac{1}{-30}}$

$\mathsf{\dfrac{2+1}{-30}= \dfrac{1}{v} }$

$\mathsf{\dfrac{3}{-30}= \dfrac{1}{v}}$

$\mathsf{-10 cm = v}$

Since, the object distance is negative it means that image is formed on same side as of object.

$\large \boxed{Magnification = \dfrac{u}{v} = \dfrac{I}{O}}$

$\mathsf{Magnification = \dfrac{10}{30} = \dfrac{I}{2}}$

$\mathsf{\dfrac{10 \times 2}{30} = I}$

$\mathsf{\dfrac{2}{3} = I}$

$\large \boxed{Image \: size = 0.66 cm}$

$\bold{Nature \: of \: image}$

▶Since, the object size is less than one, it means image is diminished.

▶Image is erect and virtual.

▶Image is formed at same side as of object.