Find the position,nature,and size of the imag | vedclass

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Question

(A) Given: Object height $h = 3 \, cm$,Object distance $u = -9 \, cm$,Focal length $f = -18 \, cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v

Vedclass · Accepted Answer

Position: $18 \, cm$ behind the mirror,Nature: Virtual and erect,Size: $6 \, cm$

Vedclass · Answer

(A) Given: Object height $h = 3 \, cm$,Object distance $u = -9 \, cm$,Focal length $f = -18 \, cm$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{-18} = \frac{1}{v} + \frac{1}{-9}$.$\frac{1}{v} = \frac{1}{9} - \frac{1}{18} = \frac{2-1}{18} = \frac{1}{18}$.Therefore,$v = +18 \, cm$. Since $v$ is positive,the image is formed $18 \, cm$ behind the mirror,which means it is virtual and erect.Now,magnification $m = \frac{h'}{h} = -\frac{v}{u}$.$h' = -\frac{v 	imes h}{u} = -\frac{18 	imes 3}{-9} = 6 \, cm$.Thus,the image is virtual,erect,$18 \, cm$ behind the mirror,and $6 \, cm$ in size.

Find the position,nature,and size of the image of an object $3 \, cm$ high placed at a distance of $9 \, cm$ from a concave mirror of focal length $18 \, cm$.

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