An object of length 6 cm is placed at a dist | vedclass

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Question

(A) Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.For a concave mirror,$f$ is negative,so $\frac{1}{-f} = \frac{1}{v} + \frac{1}

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$-2$

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(A) Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.For a concave mirror,$f$ is negative,so $\frac{1}{-f} = \frac{1}{v} + \frac{1}{-4f}$.Rearranging gives: $\frac{1}{v} = \frac{1}{4f} - \frac{1}{f} = \frac{1-4}{4f} = -\frac{3}{4f}$.Thus,$v = -\frac{4f}{3}$.The magnification $m$ is given by $m = \frac{I}{O} = -\frac{v}{u}$.Substituting the values: $\frac{I}{6} = -\frac{(-4f/3)}{-4f}$.$\frac{I}{6} = -\frac{1}{3}$.$I = -\frac{6}{3} = -2 \ cm$.The magnitude of the length of the image is $2 \ cm$ (the negative sign indicates an inverted image).

An object of length $6 \ cm$ is placed at a distance of $4f$ on the principal axis of a concave mirror of focal length $f$. The length of the image will be ......$cm$.

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