Ray Optics

JEE Main level, single correct, medium

290 of 295

Question 290 of 295

The focal length of the objective and the eyepiece of a telescope are 50 cm50 \mathrm{~cm} and 5 cm5 \mathrm{~cm} respectively. If the telescope is focussed for distinct vision on a scale distant 2 m2 \mathrm{~m} from its objective, then its magnifying power will be :

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Answer: (D) −2-2

Given: f0=50 cm,fe=5 cm\mathrm{f}_0=50 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm}
d=25 cm,u0=−200 cm\mathrm{d}=25 \mathrm{~cm}, \mathrm{u}_0=-200 \mathrm{~cm}
Magnification M=\mathrm{M}= ?
As 1v0−1u0=1f0\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0}
⇒1v0=1f0+1u0=150−1200=4−1200=3200\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}
or v0=2003 cm\mathrm{v}_0=\frac{200}{3} \mathrm{~cm}
Now ve=d=−25 cm\mathrm{v}_{\mathrm{e}}=\mathrm{d}=-25 \mathrm{~cm}
From, 1ve−1ue=1fe\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}
−1ue=1fe−1ve=15+125=625\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}
or, ve=−256 cm\mathrm{v}_{\mathrm{e}}=\frac{-25}{6} \mathrm{~cm}

 Magnification M=M0×Me=v0u0×veue=−200/3200×−25−25/6=−13×6=−2\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}

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