Question

An object of 4cm high is placed at a distance of 6cm infront of a concave mirror of focal length 12cm. Find the position, nature and size of the image formed​.
class-10
subject-Physics​

Answer 1

★Figure refer to attachment

Given

• Object height = + 4cm
• Object distance = - 6cm
• Focal length = - 12cm

Find out

• Position
• Nature
• Size of image

Solution

Apply Mirror Formula

$\implies\sf \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{v}+\dfrac{-1}{6}=\dfrac{-1}{12} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{1}{6}-\dfrac{1}{12} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{2-1}{12} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{1}{12} \\ \\ \\ \implies\sf v=+12cm$

Hence, Image distance is +12cm

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Magnification : It is the ratio of height or size of image to the height or size of object.

$\implies\sf m=\dfrac{-v}{u}=\dfrac{hi}{ho} \\ \\ \\ \implies\sf \dfrac{-v}{u}=\dfrac{hi}{ho} \\ \\ \\ \implies\sf \dfrac{-12}{-6}=\dfrac{hi}{4} \\ \\ \\ \implies\sf 2=\dfrac{hi}{4} \\ \\ \\ \implies\sf hi=+8cm$

Hence,

• Height of image = + 8cm
• Nature of image =virtual and erect

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