KCET · Physics · Atomic Physics
Diameter of the objective of a telescope is \( 200 \mathrm{~cm} \). What is the resolving power of a telescope
\( ? \) Take wavelength of light \( =5000 \mathrm{~A} \).
- A \( 6.56 \times 10^{6} \)
- B \( 3.28 \times 10^{5} \)
- C \( 1 \times 10^{6} \)
- D \( 3.28 \times 10^{6} \)
Answer & Solution
Correct Answer
(D) \( 3.28 \times 10^{6} \)
Step-by-step Solution
Detailed explanation
Resolving power of telescope is given as
\[
\begin{array}{l}
\frac{D}{1.22 \lambda}=\frac{200 \mathrm{~cm}}{1.22 \times 5000 \mathrm{AA}} \\
=\frac{200 \times 10^{-2} \mathrm{~m}}{1.22 \times 5000 \times 10^{-10} \mathrm{~m}} \\
=3.2786 \times 10^{6} \mathrm{~m}
\end{array}
\]
Therefore, resolving power of telescope \( =3.28 \times 10^{6} \)
\[
\begin{array}{l}
\frac{D}{1.22 \lambda}=\frac{200 \mathrm{~cm}}{1.22 \times 5000 \mathrm{AA}} \\
=\frac{200 \times 10^{-2} \mathrm{~m}}{1.22 \times 5000 \times 10^{-10} \mathrm{~m}} \\
=3.2786 \times 10^{6} \mathrm{~m}
\end{array}
\]
Therefore, resolving power of telescope \( =3.28 \times 10^{6} \)
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