Question

An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave
mirror of focal length 15.0 cm. At what distance from the mirror
should a screen be placed in order to obtain a sharp image? Find
the nature and the size of the image.​

Answer 1

Answer:

• The screen should be placed at 37.5 in front of the mirror.
• The size of the image is 6 cm
• The nature of the image is real, inverted, magnified

Explanation:

Given :

An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave  mirror of focal length 15.0 cm.

To find :

• the distance from the mirror  should a screen be placed in order to obtain a sharp image
• the nature and the size of the image

Solution :

size of the object, hₒ = 4 cm

distance of the object from the mirror, u = -25 cm (-ve as the object is in front of the mirror)

focal length of the mirror, f = -15 cm (-ve as the mirror is concave)

By mirror formula,

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

Substituting the values,

  $\sf \dfrac{1}{-15}=\dfrac{1}{-25}+\dfrac{1}{v} \\\\ \sf \dfrac{1}{v}=\dfrac{-1}{15}+\dfrac{1}{25} \\\\ \sf \dfrac{1}{v}=\bigg(\dfrac{-1}{15} \times \dfrac{10}{10} \bigg) +\bigg(\dfrac{1}{25} \times \dfrac{6}{6} \bigg) \\\\ \sf \dfrac{1}{v}=\dfrac{-10}{150}+\dfrac{6}{150} \\\\ \sf \dfrac{1}{v}=\dfrac{-10+6}{150} \\\\ \sf \dfrac{1}{v}=\dfrac{-4}{150} \\\\ v=\dfrac{-150}{4} \\\\ \sf v=-37.5 \ cm$

The screen should be placed at 37.5 in front of the mirror.

We also know,

 $\boxed{\sf \dfrac{h_i}{h_o}=\dfrac{-v}{u}}$

Substitute the values,

 $\sf \dfrac{h_i}{4}=\dfrac{-(-37.5)}{-25} \\\\ \sf h_i=\dfrac{-37.5}{25} \times 4 \\\\ \sf h_i=\dfrac{-150}{25} \\\\ \sf h_i=-6 \ cm$

The size of the image is 6 cm

-ve sign indicates the image is inverted.

Nature of the image :

• real

• inverted

• magnified