A straight metal rod of length 6 cm is place | vedclass

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(D) Given: Focal length $f = -9 \ cm$ (concave mirror). The rod is placed along the principal axis. The end closer to the mirror is at $u_1 = -15 \ cm

Vedclass · Accepted Answer

$6.75$

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(D) Given: Focal length $f = -9 \ cm$ (concave mirror). The rod is placed along the principal axis. The end closer to the mirror is at $u_1 = -15 \ cm$. The length of the rod is $6 \ cm$,so the farther end is at $u_2 = -15 - 6 = -21 \ cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we find the image positions $v_1$ and $v_2$.For $u_1 = -15 \ cm$: $\frac{1}{v_1} = \frac{1}{-9} - \frac{1}{-15} = \frac{-5 + 3}{45} = \frac{-2}{45} \implies v_1 = -22.5 \ cm$.For $u_2 = -21 \ cm$: $\frac{1}{v_2} = \frac{1}{-9} - \frac{1}{-21} = \frac{-7 + 3}{63} = \frac{-4}{63} \implies v_2 = -15.75 \ cm$.The length of the image is $|v_1 - v_2| = |-22.5 - (-15.75)| = |-6.75| = 6.75 \ cm$.

$A$ straight metal rod of length $6 \ cm$ is placed along the principal axis of a concave mirror of focal length $9 \ cm$ such that the end of the rod closer to the mirror is at a distance of $15 \ cm$ from the pole of the mirror. The length of the image of the rod is (in $cm$)

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Assertion : The formula connecting $u, v$ and $f$ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.
Reason : Laws of reflection are strictly valid for plane surfaces,but not for large spherical surfaces.

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