CGPET 2021 Solved Paper

According to the problem, focal length of objective, $f_o = 50\ \mathrm{cm}$ Object distance, $u_o = -200\ \mathrm{cm}$ Image distance, $v_o = ?$ Now, from lens formula, $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \Rightarrow \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o}$ ⇒ $\frac{1}{v_o} = \frac{1}{50} + \frac{1}{-200}$ ∴ $v_o = \frac{200}{3}\ \mathrm{cm}$........(i) Also, focal length of eyepiece, $f_e = 5\ \mathrm{cm},\ v_e = -D = -25\ \mathrm{cm}$ Again for eyepiece, using lens formula, $\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$ ⇒ $\frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e}$ Put the given values, we get $\frac{1}{u_e} = \frac{1}{-25} - \frac{1}{5}$ ∴$u_e = -\frac{25}{6}\ \mathrm{cm}$ .....(ii) Since, the separation between objective and eyepiece is $L = |v_o| + |u_e|$ From eqs. (i) and (ii), we get $L = \frac{200}{3} + \frac{25}{6}$ ⇒ $L = 70.83 \simeq 70.8\ \mathrm{cm}$