Question

Good Morning brainlians!!

A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus located at 5.00 m from this mirror, find the position,nature and size of the image.


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Answer 1

Answer:

$\red{\rm\underline\bold{Given}}$

• Radius of curvature, R = +3.00 m
• Object distance, u = -5.00 m

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$\red{\rm\underline\bold{To \: Find} }$

• Image distance, v = ?
• Height of the image, h' = ?

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$\red{\rm\underline\bold{Solution} }$

Focal length,

$\sf \: f = \frac{R}{2} = + \frac{3.00}{2} = 1.50 \: m$

Since,

$\sf \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\sf \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = + \frac{1}{1.50} - \frac{1}{ - 5.00}$

$\sf \frac{1}{1.50} + \frac{1}{5.00} = \frac{2}{3} + \frac{1}{5}$

$\sf \: \frac{1}{v} = \frac{13}{15}$

$\sf v = \frac{15}{13} = \bold{ + 1.15 \: m}$

The image is 1.15 m at back of the mirror.

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Magnification,

$\sf \: m = \frac{h'}{h} = - \frac{v}{u} = - \frac{1.15 \: m}{ - 5.00 \: m} = \bold{ + 0.23}$

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

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Answer 2

refers in attachment. . . . !

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