Solution:

Ans. (i) Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.

$$ m=\frac{\beta}{\alpha}=\frac{f_{0}}{f_{e}} $$

(ii) Expression :

$$ \begin{equation*} m=\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right) \tag{1} \end{equation*} $$

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Using the lens equation for an objective lens, $\mathbf{1}$

$$ \begin{aligned} \frac{1}{f_{0}} & =\frac{1}{v_{o}}-\frac{1}{u_{o}} \ \Rightarrow \quad \frac{1}{150} & =\frac{1}{v_{0}}-\frac{1}{3 \times 10^{5}} \end{aligned} $$

$$ \begin{aligned} \Rightarrow \quad \frac{1}{v_{0}} & =\frac{1}{150}-\frac{1}{3 \times 10^{5}}=\frac{2000-1}{3 \times 10^{5}} \mathrm{1} / 2 \ \Rightarrow \quad v_{\mathrm{o}} & =\frac{3 \times 10^{5}}{1999} \mathrm{~cm} \ & \approx 150 \mathrm{~cm} \end{aligned} $$

Hence, magnification due to the objective lens

$$ \begin{aligned} & m_{o}=\frac{v_{0}}{u_{0}}=\frac{150 \times 10^{-2} \mathrm{~m}}{3000 \mathrm{~m}} \ & m_{o} \approx \frac{10^{-2}}{20}=0.05 \times 10^{-2} \end{aligned} $$

Using lens formula for eyepiece,

$$ \begin{array}{rlrl} & & \frac{1}{f_{e}} & =\frac{1}{v_{e}}-\frac{1}{u_{e}} \ \Rightarrow & \frac{1}{5} & =\frac{1}{-25}-\frac{1}{u_{e}} \ \Rightarrow & \frac{1}{u_{e}} & =\frac{1}{-25}-\frac{1}{5}=\frac{-1-5}{25} \ \Rightarrow & u_{e} & =\frac{-25}{6} \mathrm{~cm} \end{array} $$

$\therefore$ Magnification due to eyepiece,

$$ m_{e}=\frac{-25}{-\frac{25}{6}}=6 $$

Hence, total magnification, $m=m_{e} \times m_{0}$

$$ m=6 \times 5 \times 10^{-4}=30 \times 10^{-4} $$

Hence, size of the final image

$$ \begin{aligned} & =30 \times 10^{-4} \times 100 \mathrm{~m} \ & =30 \mathrm{~cm} \end{aligned} $$

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