An object placed at a distance of 16 cm from | vedclass

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(A) For a convex lens, the magnification $m$ is given by $m = \frac{f}{f+u}$, where $u$ is the object distance (using sign convention, $u$ is negative

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) For a convex lens, the magnification $m$ is given by $m = \frac{f}{f+u}$, where $u$ is the object distance (using sign convention, $u$ is negative).Given that the magnification $m$ is the same for two different object positions, this is only possible if one image is real and the other is virtual.For the first case, $u_1 = -16 \ cm$. The magnification is $m = \frac{f}{f-16}$. Since $m > 1$, this must be a virtual image, so $m = \frac{f}{f-16}$.When the object is moved towards the lens by $8 \ cm$, the new object distance is $u_2 = -(16 - 8) = -8 \ cm$. The magnification is $m' = \frac{f}{f-8}$. Since the magnitude of magnification is the same, we have $|\frac{f}{f-16}| = |\frac{f}{f-8}|$.Since $m > 1$, one must be positive and one negative: $\frac{f}{f-16} = -\frac{f}{f-8}$.Dividing by $f$ (as $f 
eq 0$): $\frac{1}{f-16} = -\frac{1}{f-8}$.$\Rightarrow f - 8 = -(f - 16) = -f + 16$.$\Rightarrow 2f = 24$.$\Rightarrow f = 12 \ cm$.

An object placed at a distance of $16 \ cm$ from a convex lens produces an image of magnification $m$ $(m > 1)$. If the object is moved towards the lens by $8 \ cm$, then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is (in $cm$)

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