Question

A virtual image is formed 20 cm from a concave mirror, having a radius of curvature of 40 cm. Find the position of the object.​

Answer 1

Since, the image is virtual so the image is formed behind the mirror. Image distance v=+20cm. R=-40cm, therefore, f=-20cm.
The equation is 1/v+1/u=1/f
=>1/20cm+1/u=1/-20cm
=>1/u=-1/20cm-1/20cm
=>1/u=-1/10cm
=>u=-10cm
The object is placed 10cm in front of the mirror.
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Answer 2

Given:

• Image distance, v = 20 cm
• Radius of curvature, R = - 40 cm

To Find:

• Object distance, u = ?

Formulae used:

$\\ \bigstar{ \underline{ \boxed { \bf{ \pink{Focal \: length \: F \: = \frac{R}{2}}}}}}$

$\bigstar{ \underline{ \boxed{ \bf{ \red{Mirror \: Formula \: \frac{1}{f} = \frac{1}{v} + \frac{1}{u}}}}}}$

Solution:

★ Putting the values from the formula :

$\\ \sf\dashrightarrow f = \frac{R}{2}$

$\sf\dashrightarrow f =\frac{-40}{2} - \frac{ 40}{2}$

$\sf\dashrightarrow f = - 20 cm$

Hence, the focal length is - 20 cm.

★ calculating the object distance:

$\\ \sf\dashrightarrow \frac {1}{f} =\frac{1}{v} + \frac{1}{u}$

$\sf\dashrightarrow \frac{1}{-20} = \frac{1}{20} + \frac{1}{u}$

$\sf\dashrightarrow \frac{1}{u} =\frac{1}{20} + \frac{1}{20}$

$\sf\dashrightarrow \frac{1}{u} =1+ \frac{1}{20}$

$\sf\dashrightarrow \frac{1}{u} = {\frac{2}{20}}$

$\sf\dashrightarrow \frac{1}{u} =\frac{1}{10} + \frac{1}{u}$

$\sf\dashrightarrow u = 10 cm$

${\bigstar{\blue{\underline{\boxed {\sf{u = 10\:cm}}}}}}$

Hence, the object distance is 10 cm.