Question

An object is placed at a distance of 10 cm from a convex mirror of focal length 20 cm. Find the position and nature
of the image.​

Answer 1

ANSWER:

NATURE OF IMAGE

• VIRTUAL
• ERECT
• DIMINISHED

POSITION OF IMAGE

• 20/3 or 6.67cm behind the mirror

GIVEN:

• Object distance = (-10) cm
• Image distance = ?
• Focal length = +20cm

TO FIND:

• Position and nature of the image.

SOLUTION:

By using mirror formula

$\implies \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \implies \: \frac{1}{v} + \frac{( - 1)}{10} = \frac{1}{20} \\ \implies \: \frac{1}{v} = \frac{1}{20} + \frac{1}{10} \\ \implies \: \frac{1}{v} = \frac{3}{20} \\ \implies \: v = \frac{20}{3} cm$

Now finding magnification

$m = \frac{ - v}{u} \\ m = \frac{ - 20}{3} \times \frac{1}{ - 10} \\ m = \frac{2}{3}$

NATURE OF IMAGE

• VIRTUAL
• ERECT
• DIMINISHED

POSITION OF IMAGE

• 20/3 or 6.67 cm behind the mirror