The focal length of the objective and the eyepiece of a telescope are 50 ~cm and 5 ~cm respectively. If the telescope is focussed for distinct vision…

Given: $\mathrm{f}_0=50 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm}$f0​=50 cm,fe​=5 cm $\mathrm{d}=25 \mathrm{~cm}, \mathrm{u}_0=-200 \mathrm{~cm}$d=25 cm,u0​=−200 cm Magnification $\mathrm{M}=$M= ? As $\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0}$v0​1​−u0​1​=f0​1​ $$\Rightarrow \frac{1}{\mathrm{v}_0}=\frac{1}{\mathrm{f}_0}+\frac{1}{\mathrm{u}_0}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}$$⇒v0​1​=f0​1​+u0​1​=501​−2001​=2004−1​=2003​ or $\mathrm{v}_0=\frac{200}{3} \mathrm{~cm}$v0​=3200​ cm Now $\mathrm{v}_{\mathrm{e}}=\mathrm{d}=-25 \mathrm{~cm}$ve​=d=−25 cm From, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$ve​1​−ue​1​=fe​1​ $$\begin{aligned} & -\frac{1}{u_e}=\frac{1}{f_e}-\frac{1}{v_e} \\ & =\frac{1}{5}+\frac{1}{25}=\frac{6}{25} \end{aligned}$$​−ue​1​=fe​1​−ve​1​=51​+251​=256​​ or, $\mathrm{v}_{\mathrm{e}}=\frac{-25}{6} \mathrm{~cm}$ve​=6−25​ cm $$\begin{aligned} & \text { Magnification } \mathrm{M}=\mathrm{M}_0 \times \mathrm{M}_{\mathrm{e}} \\ & =\frac{\mathrm{v}_0}{\mathrm{u}_0} \times \frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{u}_{\mathrm{e}}}=\frac{-200 / 3}{200} \times \frac{-25}{-25 / 6} \\ & =-\frac{1}{3} \times 6=-2 \end{aligned}$$​ Magnification M=M0​×Me​=u0​v0​​×ue​ve​​=200−200/3​×−25/6−25​=−31​×6=−2​