A convex lens with lateral magnification 2 is | vedclass

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(B) Let $u_1$ be the initial object distance from the lens and $f$ be the focal length of the lens.Given: $v_1 = 60 \ cm$ (initial image distance) and

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(B) Let $u_1$ be the initial object distance from the lens and $f$ be the focal length of the lens.Given: $v_1 = 60 \ cm$ (initial image distance) and $m = 2$ (lateral magnification).Using the magnification formula $m = \frac{v_1}{u_1} = 2$ (for a real image formed by a convex lens, $v$ is positive and $u$ is negative, so $m = v/u$ in magnitude for the distance ratio), we have:$\frac{60}{|u_1|} = 2 \Rightarrow |u_1| = 30 \ cm$. Thus, $u_1 = -30 \ cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1}$:$\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20} \Rightarrow f = 20 \ cm$.When liquid of height $H = 24 \ cm$ is filled, the object appears to shift upwards due to refraction. The new apparent object distance $u_2$ is related to the original distance by the normal shift formula: $u_2 = |u_1| - H(1 - \frac{1}{\mu})$.Using the lens formula for the new image distance $v_2 = 120 \ cm$:$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \Rightarrow \frac{1}{20} = \frac{1}{120} - \frac{1}{u_2} \Rightarrow \frac{1}{u_2} = \frac{1}{120} - \frac{1}{20} = \frac{1-6}{120} = -\frac{5}{120} = -\frac{1}{24}$.So, $|u_2| = 24 \ cm$.Equating the shift: $|u_1| - |u_2| = H(1 - \frac{1}{\mu})$$30 - 24 = 24(1 - \frac{1}{\mu}) \Rightarrow 6 = 24(1 - \frac{1}{\mu})$$1 - \frac{1}{\mu} = \frac{6}{24} = \frac{1}{4} \Rightarrow \frac{1}{\mu} = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow \mu = \frac{4}{3} \approx 1.33$.

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