Question

Arrange the following telescopes, where \( D \) is the telescope diameter and \( \lambda \) is the wavelength, in order of decreasing resolving power:
I. \( D = 100\,\text{m}, \lambda = 21\,\text{cm} \)
II. \( D = 2\,\text{m}, \lambda = 500\,\text{nm} \)
III. \( D = 1\,\text{m}, \lambda = 100\,\text{nm} \)
IV. \( D = 2\,\text{m}, \lambda = 10\,\text{mm} \)

• A. III, II, IV, I
• B. II, III, I, IV
• C. IV, III, II, I
• D. III, II, I, IV

Hint:

Resolving power of a telescope increases with larger aperture size and decreases with higher wavelength. Always compare \( \frac{D}{\lambda} \) to rank resolving power.

Answer 1

The Correct Option is D

Step 1: Recall the formula for resolving power.
The resolving power \( R \) of a telescope is given by \[ R \propto \frac{D}{\lambda}, \] where \( D \) is the aperture diameter and \( \lambda \) is the wavelength. Higher \( D \) and lower \( \lambda \) give better resolution.
Step 2: Calculate relative values of \( \frac{D}{\lambda} \) for each case.
\[ \text{I. } \frac{D}{\lambda} = \frac{100}{0.21} = 476.19 \] \[ \text{II. } \frac{D}{\lambda} = \frac{2}{5 \times 10^{-7}} = 4.0 \times 10^{6} \] \[ \text{III. } \frac{D}{\lambda} = \frac{1}{1 \times 10^{-7}} = 1.0 \times 10^{7} \] \[ \text{IV. } \frac{D}{\lambda} = \frac{2}{1 \times 10^{-2}} = 200 \]
Step 3: Arrange in decreasing order of resolving power.
\[ III>II>I>IV \]
Step 4: Final Answer.
Therefore, the order of decreasing resolving power is III, II, I, IV.