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The distance of the image from the lens V=22.5cm and the height of the image h=6.25cm.
It is given that
Hieght of an object H = 5cm
Distance of an object =18cm
Focal lenght of lens = 10cm
As we know that the lens formula is given by
[tex]\dfrac{1}{f} =\dfrac{1}{v} +\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v} =\dfrac{1}{10} -\dfrac{1}{18}=\dfrac{8}{180}[/tex]
so now
[tex]v= \dfrac{180}{8} =22.5\ cm[/tex]
Now to find the height of the image
[tex]\rm magnification =\dfrac{v}{u} =\dfrac{-Image\ Hieght }{object\ height}[/tex]
[tex]\rm magnification =\dfrac{22.5}{18} =\dfrac{-x}{5}[/tex]
[tex]x=-6.25\ cm[/tex]
Thus the distance of the image from the lens is V=22.5cm and the height of the image h=6.25cm.
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