The magnification produced by a astronomical telescope for normal...

Given, $m=10$, length of telescope $=1.1 \mathrm{ m} We know that, Magnification,$m=$\frac{t_0}{t_e}$$\begin{aligned} & 10=\frac{t_0}{t_e} \\ & f_0=10 f_e \\ & \because \quad f_e+f_0=1.1 \mathrm{ m} \\ & \Rightarrow \quad t_e+10 t_e=1.1 \mathrm{ m} \\ & {\left[\because f_0=10 f_e\right]} \\ & \Rightarrow \quad f_e(1+10)=1.1 \mathrm{ m} \\ & \Rightarrow \quad f_e=0.1 \mathrm{ m} \text{\text{ or } } 10 \mathrm{ cm} \\ & \end{aligned}$$$Magnification least distance of distinct vision,$$\begin{aligned} M_b & =\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right) \\ & =10\left(1+\frac{10}{25}\right) \quad[\because D=25 \mathrm{ cm}] \\ & =10 \times \frac{35}{25} \\ & =14\end{aligned}$$