Question

(8. Where should an object be placed before a concave
mirror of focal length 20 cm so that a real image is
formed at a distance of 60 cm from it?​

Answer 1

Answer :-

u = -20 cm

Given :-

f = - 20 cm

v = 60 cm

To find :-

The placed where the object is placed or object distance.

Solutions:-

Let the object distance be u.

• By using mirror formula,

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

• Put the given values,

→$\mathsf{\dfrac{1}{(-20)} = \dfrac{1}{u} + \dfrac{1}{60}}$

→$\mathsf{\dfrac{1}{u} = \dfrac{-1}{20}-\dfrac{1}{60}}$

→$\mathsf{\dfrac{1}{u} = \dfrac{-3-1}{60}}$

→$\mathsf{\dfrac{1}{u}=\dfrac{-4}{60}}$

→$\mathsf{u = \dfrac{-60}{4}}$

→$\mathsf{u = -20 cm}$

hence,

The object is placed at a distance of 20 cm from the concave mirror.

Answer 2

Answer:-

According to the given question:-

• 20cm is the (f)
• 60cm KS the (v)
• Let (x) be the object distance.

Mirror formula:-

$= > \: \frac{1}{f} = \frac{1}{x} + \frac{1}{v} (mirror \: formula)$

Adding values:-

$= > \: ( \frac{1}{20} = \frac{1}{x} + \frac{1}{16} )$

$= > ( \frac{1}{x} = \frac{ - 1}{20} - \frac{1}{60} )$

$= > ( \frac{1}{x} = \frac{ - 3 - 1}{60} )$

$= > ( \frac{1}{x} = \frac{ - 4}{60})$

$= > (x = \frac{ - 60}{4} )$

$= > (x = - 20cm)$

Therefore, 20 CM AI the distance of object from the concave mirror.