An object 1 , cm tall is placed at a distance | vedclass

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Question

(A) Given: Object height $O = 1 \, cm$,Object distance $u = -15 \, cm$,Focal length $f = -10 \, cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v

Vedclass · Accepted Answer

$v = -30 \, cm$,$I = -2 \, cm$,Real and Inverted

Vedclass · Answer

(A) Given: Object height $O = 1 \, cm$,Object distance $u = -15 \, cm$,Focal length $f = -10 \, cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = \frac{-3 + 2}{30} = \frac{-1}{30}$.Thus,$v = -30 \, cm$. The negative sign indicates the image is formed in front of the mirror.Magnification $m = \frac{I}{O} = -\frac{v}{u}$.$I = -\frac{v}{u} 	imes O = -\frac{-30}{-15} 	imes 1 = -2 \, cm$.The image is $30 \, cm$ in front of the mirror,$2 \, cm$ tall,and is real and inverted.

An object $1 \, cm$ tall is placed at a distance of $15 \, cm$ from a concave mirror of focal length $10 \, cm$. Find the position,size,and nature of the image formed.

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