Consider a concave mirror of 10 ,cm focal len | vedclass

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Question

(A) Using the mirror equation,we have:$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$Given that for a concave mirror,the focal length $f = -10 \,cm$ and the

Vedclass · Accepted Answer

$-16.7 \,cm$ and $-0.67$

Vedclass · Answer

(A) Using the mirror equation,we have:$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$Given that for a concave mirror,the focal length $f = -10 \,cm$ and the object distance $u = -25 \,cm$.Substituting these values into the equation:$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-10} - \frac{1}{-25}$$\frac{1}{v} = -\frac{1}{10} + \frac{1}{25} = \frac{-5 + 2}{50} = -\frac{3}{50}$$v = -\frac{50}{3} \approx -16.7 \,cm$The magnification $m$ is given by:$m = -\frac{v}{u} = -\frac{-16.7}{-25} = -0.67$Thus,the image is formed at $-16.7 \,cm$ and the magnification is $-0.67$.

Consider a concave mirror of $10 \,cm$ focal length illuminated by an object kept at a distance of $25 \,cm$. The distance at which the image is formed and its magnification respectively are

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