Question

A convex mirror used for rear view on an automobile has a radius of curvater of 3.00metre if a bus is located at 5.00metre from this mirror find the position nature and size of the image

Answer 1

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• Radius of curvature (R) = 3.00 m
• object distance (u) = - 5.00 m

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• Image distance (v) = ?
• size of the image =?

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we know that,

$\bf\:F=\dfrac{R}{2}$

$\bf\:F=\dfrac{3.00}{2}$

$\bf\:F=1.50\:m$

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

$\bf\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\bf\dfrac{1}{v}=\dfrac{1}{1.50}-\dfrac{1}{-5.00}$

$\bf\dfrac{1}{v}=\dfrac{1}{1.50}+\dfrac{1}{5}$

$\bf\dfrac{1}{v}=\dfrac{10}{15}+\dfrac{1}{5}$

$\bf\dfrac{1}{v}=\dfrac{10+3}{15}$

$\bf\dfrac{1}{v}=\dfrac{13}{15}$

$\bf\:v=1.15\:m$

Therefore image is produced at a distance of 1.15m

Finding size of image ,

Magnification=$\bf\dfrac{-v}{u}$

$\bf\:=\dfrac{-Image\: distance}{Object\:distance}$

$\bf\:=\dfrac{-1.15}{-5}$

$\bf\:=\dfrac{115}{500}$

$\bf\:=\dfrac{23}{100}$

$\bf\:=0.23$

The image is virtual,erect and smaller in size by a factor of 0.23