Search any question & find its solution
Question:
Answered & Verified by Expert
The focal length of the objective and the eyepiece of a telescope are $50 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the telescope is focussed for distinct vision on a scale distant $2 \mathrm{~m}$ from its objective, then its magnifying power will be :
Options:
Solution:
2117 Upvotes
Verified Answer
The correct answer is:
$-2$
$-2$
Given: $\mathrm{f}_0=50 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm}$
$\mathrm{d}=25 \mathrm{~cm}, \mathrm{u}_0=-200 \mathrm{~cm}$
Magnification $\mathrm{M}=$ ?
As $\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0}$
$$
\Rightarrow \frac{1}{\mathrm{v}_0}=\frac{1}{\mathrm{f}_0}+\frac{1}{\mathrm{u}_0}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}
$$
or $\mathrm{v}_0=\frac{200}{3} \mathrm{~cm}$
Now $\mathrm{v}_{\mathrm{e}}=\mathrm{d}=-25 \mathrm{~cm}$
From, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$$
\begin{aligned}
& -\frac{1}{u_e}=\frac{1}{f_e}-\frac{1}{v_e} \\
& =\frac{1}{5}+\frac{1}{25}=\frac{6}{25}
\end{aligned}
$$
or, $\mathrm{v}_{\mathrm{e}}=\frac{-25}{6} \mathrm{~cm}$
$$
\begin{aligned}
& \text { Magnification } \mathrm{M}=\mathrm{M}_0 \times \mathrm{M}_{\mathrm{e}} \\
& =\frac{\mathrm{v}_0}{\mathrm{u}_0} \times \frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{u}_{\mathrm{e}}}=\frac{-200 / 3}{200} \times \frac{-25}{-25 / 6} \\
& =-\frac{1}{3} \times 6=-2
\end{aligned}
$$
$\mathrm{d}=25 \mathrm{~cm}, \mathrm{u}_0=-200 \mathrm{~cm}$
Magnification $\mathrm{M}=$ ?
As $\frac{1}{\mathrm{v}_0}-\frac{1}{\mathrm{u}_0}=\frac{1}{\mathrm{f}_0}$
$$
\Rightarrow \frac{1}{\mathrm{v}_0}=\frac{1}{\mathrm{f}_0}+\frac{1}{\mathrm{u}_0}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}
$$
or $\mathrm{v}_0=\frac{200}{3} \mathrm{~cm}$
Now $\mathrm{v}_{\mathrm{e}}=\mathrm{d}=-25 \mathrm{~cm}$
From, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$
$$
\begin{aligned}
& -\frac{1}{u_e}=\frac{1}{f_e}-\frac{1}{v_e} \\
& =\frac{1}{5}+\frac{1}{25}=\frac{6}{25}
\end{aligned}
$$
or, $\mathrm{v}_{\mathrm{e}}=\frac{-25}{6} \mathrm{~cm}$
$$
\begin{aligned}
& \text { Magnification } \mathrm{M}=\mathrm{M}_0 \times \mathrm{M}_{\mathrm{e}} \\
& =\frac{\mathrm{v}_0}{\mathrm{u}_0} \times \frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{u}_{\mathrm{e}}}=\frac{-200 / 3}{200} \times \frac{-25}{-25 / 6} \\
& =-\frac{1}{3} \times 6=-2
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.