A 4.5, cm needle is placed 12, cm away from a | vedclass

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Question

(A) Given: Object height $O = 4.5\, cm$,Object distance $u = -12\, cm$,Focal length $f = +15\, cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u

Vedclass · Accepted Answer

Location: $6.7\, cm$ behind the mirror,Magnification: $0.56$

Vedclass · Answer

(A) Given: Object height $O = 4.5\, cm$,Object distance $u = -12\, cm$,Focal length $f = +15\, cm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{v} + \frac{1}{-12} = \frac{1}{15}$.$\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60}$.$v = \frac{60}{9} \approx 6.67\, cm \approx 6.7\, cm$.The positive sign indicates that the image is formed $6.7\, cm$ behind the mirror.Magnification $m = -\frac{v}{u} = -\frac{6.67}{-12} \approx 0.56$.Since $m$ is positive and less than $1$,the image is virtual,erect,and diminished.

$A$ $4.5\, cm$ needle is placed $12\, cm$ away from a convex mirror of focal length $15\, cm$. Give the location of the image and its magnification.

Similar Questions

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$A$ $4.5 \, cm$ needle is placed $12 \, cm$ away from a convex mirror of focal length $15 \, cm$. Give the location of the image and the magnification. Describe what happens to the image as the needle is moved farther from the mirror.

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