Question

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror
of

radius
of curvature
36 cm. At what distance from the mirror
should a screen be

placed in
order to obtain a sharp image?​

Answer 1

Answer:

• $-54\:cm$

Explanation:

Given

• A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm.

Tofind

• At what distance from the mirror should a screen be placed in order to obtain a sharp image?

$\underline{\rm{From\:the\:mirror \: formula}}$

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

$\implies\: \:\: \sf{ \dfrac{1}{( - \frac{36}{2} )} = \dfrac{1}{v} + \dfrac{1}{ ( - 27)}}$

$\implies\: \:\:\sf{ \dfrac{1}{ - 18} = \dfrac{1}{v} - \dfrac{1}{27} }$

$\implies \: \:\:\sf{ \dfrac{1}{v} = - \dfrac{1}{18} + \dfrac{1}{27}}$

$\implies\: \:\: \sf{ \dfrac{1}{v} = \dfrac{ - 3 + 2}{54} }$

$\implies \:\:\: \sf{ \dfrac{1}{v} = \dfrac{ - 1}{54}}$

$\implies \: \:\:\sf{v = - 54 \: cm}$

Answer 2

Given:

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature

36 cm.

To find:

Distance at which a screen should be placed for a sharp image?

Calculation:

First, we need to check if a real image will be formed or not !

Applying MIRROR FORMULA:

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

$\implies \dfrac{1}{( - \frac{36}{2} )} = \dfrac{1}{v} + \dfrac{1}{ ( - 27)}$

$\implies \dfrac{1}{ - 18} = \dfrac{1}{v} - \dfrac{1}{27}$

$\implies \: \dfrac{1}{v} = - \dfrac{1}{18} + \dfrac{1}{27}$

$\implies \: \dfrac{1}{v} = \dfrac{ - 3 + 2}{54}$

$\implies \: \dfrac{1}{v} = \dfrac{ - 1}{54}$

$\implies \: v = - 54 \: cm$

• Since value of v is negative, real image is produced !

Hence, screen has to placed 54 cm in front of the mirror for a sharp image.