(a) (i) Ray diagram of compound microscope: In a compound microscope there are two lenses - objective $(O)$ and Eyepiece $(E)$. Object $P Q$ is placed in front of the objective at a distance more than the focal length of the objective. An inverted, magnified, real image $P_1 Q_1$ is formed in front of eyepiece within the optical centre and the focus of the eyepiece. This acts as the object of the eyepiece. An (erect with respect to $P_1 Q_1$, inverted with respect to $P Q$ ), magnified, virtual image $P_2 Q_2$ is formed at a distance $D$ (minimum distance of distinct vision) from the eyepiece. Magnification: For objective: Object distance $=u$ Image distance $=v$ $\;\text{ Magnification }\;=m_{o}=\frac{v}{u}=\frac{P_1 Q_1}{P Q}$ Applying the lens formula, $\;\frac{1}{v}-\;\frac{1}{-u}=\;\frac{1}{f_0}$ Or, $1+\;\frac{v}{u}\;=\;\frac{v}{f_0}$ Or, $\;\frac{v}{u}\;=\;\frac{v}{f_0}-1$ For eyepiece: $\;\text{ Magnification }\;=m_e=1+\;\frac{D}{f_e}$ Magnification of the combination of objective and eyepiece $=m=m_O\times m_e$ $\;\text{ Or, }\;\;\;$$\;m=\;\frac{v}{u}\times\left(1+\;\frac{D}{f e}\right)$ $\;\text{ Or, }\;\;m=\left(\;\frac{v}{f_0}-1\right)\left(1+\;\frac{D}{f e}\right)$ $P_1 Q_1$ image is formed very close to the eyepiece, hence $v$ can be approximated as the distance between the two lenses i.e., the length of the tube(L). $\therefore\;\;m=\left(\;\frac{L}{f_0}-1\right)\left(1+\;\frac{D}{f_e}\right)$ Since, $f_e\ll D$ and $f_0\ll L$, hence the above expression may be approximated as, $m=\;\frac{L}{f_0}\times\;\frac{D}{f_e}$ (ii) Given, $u=1.5\ \mathrm{cm}, f_0=1.25\ \mathrm{cm}, f_e=5\ \mathrm{cm}, D$ $=25\ \mathrm{cm}$ Here, all alphabets are in their usual meanings Applying lens formula for objective lens, $\;\frac{1}{v}-\;\frac{1}{u}\;=\;\frac{1}{f_0}$ $\;\text{ Or, }\;\;\;\frac{1}{v}-\;\frac{1}{-1.5}\;=\;\frac{1}{1.25}$ $\therefore\;v\;=7.5\ \mathrm{cm}$ Magnification $=m=\;\frac{v}{u}\times\left(1+\;\frac{D}{f_e}\right)$ $\;\;\text{ Or, }\;\;m=\;\frac{7.5}{-1.5}\times\left(1+\;\frac{25}{5}\right)$ $\therefore\;\;m\;=-30$ OR (b) (i) Astronomical telescope in normal adjustment: In an astronomical telescope there are two lenses - objective $(O)$ and Eyepiece $(E)$.The two lenses are so placed during focussing that the foci of the lenses meet at a point.Objective is directed towards the object at infinity.Parallel rays coming from the object meet at the focus of the objective and forms an inverted, real image $P_1 Q_1$ in front of eyepiece. This point is the focus of eyepiece too.This acts as the object of the eyepiece. An (inverted with respect to $P_1 Q_1$, erect with respect to original object), highly magnified, real image is formed at infinity.Magnification $=m$$=\frac{\;\text{ Angle subtended at eye by the final image }\;}{\;\text{ Angle subtended at eye by the object }\;}$ Or, $m=$ Or, $m\;=\;\frac{\beta}{\alpha}$ $\;\text{ Or, }\;\;m=\;\frac{\angle Q_1 E P_1}{\angle Q_1 O P_1}$ $\;\text{ Or, }\;\;m=\;\frac{\tan\angle Q_1 E P_1}{\tan\angle Q_1 O P_1}$ [ $\alpha$ and $\beta$ being very small, $\tan\alpha=\alpha$ and $\tan\beta=\beta$ ] Or, $\;m=\;\frac{\;\frac{Q_1 P_1}{Q_1 E}}{\frac{Q_1 P_1}{Q_1 O}}$ Or, $\;m=\;\frac{Q_1 O}{Q_1 E}$ ∴ $\;m=\;\frac{f_o}{f_e}$ $\;\;\text{ (ii) Since, }\;\;\;m=\;\frac{f_0}{f_e}$$\;\;\text{ Or, }\;\;2.9=\;\frac{f_0}{f_e}$$\;\therefore\;\;f_o=2.9 f_e$$\;\;\text{ Also, }\;\;\;f_o+f_e=150$$\;\;\text{ Or, }\;2.9 f_e+f_e=150$$\;\therefore\;\;f_e=\;\frac{150}{3.9}$$\;=38.46\ \mathrm{cm}$$\;f_O=2.9 f_c$$\;=2.9\times 38.46$$\;=111.54\ \mathrm{cm}$$\;$