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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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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:
[tex] \red{\rm\underline\bold{Given}}[/tex]
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[tex] \red{\rm\underline\bold{To \: Find} }[/tex]
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[tex] \red{\rm\underline\bold{Solution} }[/tex]
Focal length,
[tex]\sf \: f = \frac{R}{2} = + \frac{3.00}{2} = 1.50 \: m \\[/tex]
Since,
[tex]\sf \: \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\[/tex]
[tex]\sf \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = + \frac{1}{1.50} - \frac{1}{ - 5.00} \\ [/tex]
[tex]\sf \frac{1}{1.50} + \frac{1}{5.00} = \frac{2}{3} + \frac{1}{5} \\ [/tex]
[tex]\sf \: \frac{1}{v} = \frac{13}{15} \\ [/tex]
[tex]\sf v = \frac{15}{13} = \bold{ + 1.15 \: m} \\ [/tex]
The image is 1.15 m at back of the mirror.
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Magnification,
[tex]\sf \: m = \frac{h'}{h} = - \frac{v}{u} = - \frac{1.15 \: m}{ - 5.00 \: m} = \bold{ + 0.23} \\[/tex]
The image is virtual, erect and smaller in size by factor of 0.23.
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refers in attachment. . . . !
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