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(A) Let $u_1$ be the object distance from the objective lens. Given $f_o = 1 \, cm$. The image distance $v_1$ from the objective lens is given by the

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) Let $u_1$ be the object distance from the objective lens. Given $f_o = 1 \, cm$. The image distance $v_1$ from the objective lens is given by the lens formula: $\frac{1}{v_1} - \frac{1}{-u_1} = \frac{1}{f_o} \Rightarrow \frac{1}{v_1} = 1 - \frac{1}{u_1} = \frac{u_1 - 1}{u_1} \Rightarrow v_1 = \frac{u_1}{u_1 - 1}$.The magnification of the objective lens is $m_o = \frac{v_1}{u_1} = \frac{1}{u_1 - 1}$.The tube length $L = v_1 + |u_e| = 20 \, cm$,where $u_e$ is the object distance for the eye-piece. Thus,$|u_e| = 20 - v_1 = 20 - \frac{u_1}{u_1 - 1} = \frac{20u_1 - 20 - u_1}{u_1 - 1} = \frac{19u_1 - 20}{u_1 - 1}$.The magnification of the eye-piece is $m_e = \frac{D}{|u_e|} = \frac{25}{|u_e|} = \frac{25(u_1 - 1)}{19u_1 - 20}$.Total magnification $M = m_o 	imes m_e = 100 \Rightarrow \left(\frac{1}{u_1 - 1}ight) 	imes \left(\frac{25(u_1 - 1)}{19u_1 - 20}ight) = 100$.$\frac{25}{19u_1 - 20} = 100 \Rightarrow 19u_1 - 20 = 0.25 \Rightarrow 19u_1 = 20.25 \Rightarrow u_1 = \frac{20.25}{19} \approx 1.0658 \, cm$.Now,$|u_e| = 20 - v_1 = 20 - \frac{1.0658}{1.0658 - 1} = 20 - \frac{1.0658}{0.0658} \approx 20 - 16.2 = 3.8 \, cm$.Using the lens formula for the eye-piece: $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \Rightarrow \frac{1}{-25} - \frac{1}{-3.8} = \frac{1}{f_e} \Rightarrow \frac{1}{f_e} = \frac{1}{3.8} - \frac{1}{25} \approx 0.263 - 0.04 = 0.223$.$f_e \approx \frac{1}{0.223} \approx 4.48 \, cm$. The closest option is $4.5 \, cm$.

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