Search any question & find its solution
Question:
Answered & Verified by Expert
(a) A giant refracting telescope at an observatory has an objective lens of focal length $15 \mathrm{~m}$. If an eyepiece of focal length $1.0 \mathrm{~cm}$ is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^6 \mathrm{~m}$, and the radius of lunar orbit is $3.8$ $\times 10^8 \mathrm{~m}$.
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is $3.48 \times 10^6 \mathrm{~m}$, and the radius of lunar orbit is $3.8$ $\times 10^8 \mathrm{~m}$.
Solution:
2167 Upvotes
Verified Answer
Here, $\mathrm{f}_0=15 \mathrm{~m}, \mathrm{f}_{\mathrm{e}}=1.0 \mathrm{~cm}=10^{-2} \mathrm{~m}$.
(a) Angular magnification
$$
=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{10^{-2}}=15 \times 10^2=1500
$$
(b) Angle subtended by the diameter of the moon
$$
=\frac{\text { arc }}{\text { radius }}=\frac{3.48 \times 10^6}{3.8 \times 10^8}
$$
Angle subtended by image
$$
\begin{aligned}
&=\frac{\text { Arc (diameter of image) }}{\text { radius (focal length of objective) }}=\frac{\mathrm{d}}{\mathrm{f}_0}=\frac{\mathrm{d}}{15} \\
&\therefore \frac{\mathrm{d}}{15}=\frac{3.48 \times 10^6}{3.8 \times 10^8} \Rightarrow \mathrm{d}=\frac{3.48 \times 15}{3.8 \times 10^2} \\
&=13.73 \times 10^{-2} \mathrm{~m}=13.7 \mathrm{~cm} .
\end{aligned}
$$
(a) Angular magnification
$$
=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}=\frac{15}{10^{-2}}=15 \times 10^2=1500
$$
(b) Angle subtended by the diameter of the moon
$$
=\frac{\text { arc }}{\text { radius }}=\frac{3.48 \times 10^6}{3.8 \times 10^8}
$$
Angle subtended by image
$$
\begin{aligned}
&=\frac{\text { Arc (diameter of image) }}{\text { radius (focal length of objective) }}=\frac{\mathrm{d}}{\mathrm{f}_0}=\frac{\mathrm{d}}{15} \\
&\therefore \frac{\mathrm{d}}{15}=\frac{3.48 \times 10^6}{3.8 \times 10^8} \Rightarrow \mathrm{d}=\frac{3.48 \times 15}{3.8 \times 10^2} \\
&=13.73 \times 10^{-2} \mathrm{~m}=13.7 \mathrm{~cm} .
\end{aligned}
$$
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.