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(D) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. Given $f = -1\, m$ (using sign convention,$f$ is negative fo

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$0.25\, m, 1\, m, 0.5\, m$

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(D) For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. Given $f = -1\, m$ (using sign convention,$f$ is negative for a concave mirror). Let the distances be $u_P = -3\, m$ and $u_Q = -5\, m$.For face $P$:$\frac{1}{v_P} + \frac{1}{-3} = \frac{1}{-1} \implies \frac{1}{v_P} = -1 + \frac{1}{3} = -\frac{2}{3} \implies v_P = -1.5\, m$.The magnitude of the image distance is $1.5\, m$.For face $Q$:$\frac{1}{v_Q} + \frac{1}{-5} = \frac{1}{-1} \implies \frac{1}{v_Q} = -1 + \frac{1}{5} = -\frac{4}{5} \implies v_Q = -1.25\, m$.The magnitude of the image distance is $1.25\, m$.The distance between the images is $|v_P - v_Q| = |1.5 - 1.25| = 0.25\, m$.Magnification $m = -\frac{v}{u}$.For face $P$: $m_P = -\frac{-1.5}{-3} = -0.5$. Height of image $P = |m_P| 	imes 2\, m = 0.5 	imes 2 = 1\, m$.For face $Q$: $m_Q = -\frac{-1.25}{-5} = -0.25$. Height of image $Q = |m_Q| 	imes 2\, m = 0.25 	imes 2 = 0.5\, m$.Thus,the values are $0.25\, m, 1\, m, 0.5\, m$.

$A$ cube of side $2\, m$ is placed in front of a concave mirror of focal length $1\, m$ with its face $P$ at a distance of $3\, m$ and face $Q$ at a distance of $5\, m$ from the mirror. The distance between the images of face $P$ and $Q$ and the heights of the images of $P$ and $Q$ are:

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