An object of size 7.0, cm is placed at 27, cm | vedclass

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Question

(A) Object distance,$u = -27\, cm$.Object height,$h = 7\, cm$.Focal length,$f = -18\, cm$.According to the mirror formula,$\frac{1}{v} + \frac{1}{u} =

Vedclass · Accepted Answer

Distance: $54\, cm$,Size: $14\, cm$,Nature: Real and Inverted

Vedclass · Answer

(A) Object distance,$u = -27\, cm$.Object height,$h = 7\, cm$.Focal length,$f = -18\, cm$.According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = \frac{-3 + 2}{54} = \frac{-1}{54}$.Thus,$v = -54\, cm$.The screen should be placed at a distance of $54\, cm$ in front of the mirror.Magnification,$m = -\frac{v}{u} = -\frac{-54}{-27} = -2$.The negative sign of magnification indicates that the image is real and inverted.Magnification,$m = \frac{h'}{h} \implies h' = m 	imes h = -2 	imes 7 = -14\, cm$.The size of the image is $14\, cm$.

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Kerosene	$1.44$
Turpentine oil	$1.47$

An object of size $7.0\, cm$ is placed at $27\, cm$ in front of a concave mirror of focal length $18\, cm$. At what distance from the mirror should a screen be placed,so that a sharp focussed image can be obtained? Find the size and the nature of the image.

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