An object placed in front of a concave mirror | vedclass

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(D) For a concave mirror, the magnification $m$ for a real image is negative, so $m = -3$ and $m = -2$.Using the mirror formula $\frac{1}{v} + \frac{1

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$30$

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(D) For a concave mirror, the magnification $m$ for a real image is negative, so $m = -3$ and $m = -2$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and magnification $m = -\frac{v}{u}$, we have $v = -mu$.Case $1$: $u_1 = -x$, $m_1 = -3$, so $v_1 = -(-3)(-x) = -3x$.$\frac{1}{-3x} + \frac{1}{-x} = \frac{1}{f} \implies \frac{-1-3}{3x} = \frac{1}{f} \implies \frac{1}{f} = \frac{-4}{3x} \quad (i)$Case $2$: $u_2 = -(x+5)$, $m_2 = -2$, so $v_2 = -(-2)(-(x+5)) = -2(x+5)$.$\frac{1}{-2(x+5)} + \frac{1}{-(x+5)} = \frac{1}{f} \implies \frac{-1-2}{2(x+5)} = \frac{1}{f} \implies \frac{1}{f} = \frac{-3}{2(x+5)} \quad (ii)$Equating $(i)$ and $(ii)$:$\frac{-4}{3x} = \frac{-3}{2(x+5)} \implies 8(x+5) = 9x \implies 8x + 40 = 9x \implies x = 40 \ cm$.Substituting $x = 40$ into $(i)$:$\frac{1}{f} = \frac{-4}{3(40)} = \frac{-4}{120} = \frac{-1}{30} \implies f = -30 \ cm$.The magnitude of the focal length is $30 \ cm$.

An object placed in front of a concave mirror at a distance of $x \ cm$ from the pole gives a $3$ times magnified real image. If it is moved to a distance of $(x+5) \ cm$, the magnification of the image becomes $2$. The focal length of the mirror is: (in $cm$)

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