Question

An object is placed at a distance of 50 cm in front of a convex mirror of focal length 25 cm .Find the omage .What will be its nature?​

Answer 1

Explanation:

F =25cm

u = -50cm

From the mirror formula,

1/v+1/u = 1/f

1/v+(-1/50) = 1/25

1/v = 1/25+1/50

1/v = 2+1/50

1/v = 3/50

v = 50/3

v = 16.66cm

Nature of image :- virtual&erect

Answer 2

$\Large{\underline{\underline{\tt{\pink{Given}}}}}$

• In convex mirror
• Object distance = - 50cm
• Focal length = + 25cm

$\Large{\underline{\underline{\tt{\pink{Find\:out}}}}}$

• Image distance
• Nature of image

$\Large{\underline{\underline{\tt{\pink{Solution}}}}}$

Apply mirror formula

$\implies\sf \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{v}+\dfrac{(-1)}{50}=\dfrac{1}{25} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{1}{50}+\dfrac{1}{25} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{1+2}{50} \\ \\ \\ \implies\sf \dfrac{1}{v}=\cancel\dfrac{3}{50} \\ \\ \\ \implies\sf v = +16.6cm$

Hence, image distance is + 16.6cm

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• Nature of image
• Nature = Virtual and erect

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