A telescope uses light having wavelength 5000 Å and using lenses of focal length 2.5cm and 30cm. If the diameter of the aperture of the objective is 10 cm, then the resolving limit and magnifying power of the telescope is respectively

Options

A. `6.1xx10^(-6)` rad and 12
B. `5.0 xx10^(-6)` rad and 12
C. `6.1xx10^(-6)` rad and `8.3xx10^(-2)`
D. `5.0xx10^(-6)` rad and `8.3xx10^(-2)`

Answer

To solve the problem, we need to calculate two things: the resolving limit and the magnifying power of the telescope.

### Step 1: Calculate the Resolving Limit
The resolving limit (angular resolution) of a telescope can be calculated using the formula:
\[ \theta = \frac{1.22 \lambda}{D} \]
where:
- \(\theta\) is the resolving limit in radians,
- \(\lambda\) is the wavelength of light used (in meters),
- \(D\) is the diameter of the aperture (in meters).

Given:
- Wavelength \(\lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m}\)
- Diameter of the aperture \(D = 10 \, \text{cm} = 0.1 \, \text{m}\)

Substituting the values into the formula:
\[ \theta = \frac{1.22 \times (5 \times 10^{-7})}{0.1} \]
Calculating this gives:
\[ \theta = \frac{1.22 \times 5 \times 10^{-7}}{0.1} = \frac{6.1 \times 10^{-7}}{0.1} = 6.1 \times 10^{-6} \, \text{radians} \]

### Step 2: Calculate the Magnifying Power
The magnifying power \(M\) of a telescope is given by the ratio of the focal length of the objective lens \(f_o\) to the focal length of the eyepiece lens \(f_e\):
\[ M = \frac{f_o}{f_e} \]

Given:
- Focal length of the objective lens \(f_o = 30 \, \text{cm} = 0.3 \, \text{m}\)
- Focal length of the eyepiece lens \(f_e = 2.5 \, \text{cm} = 0.025 \, \text{m}\)

Substituting the values into the formula:
\[ M = \frac{0.3}{0.025} \]
Calculating this gives:
\[ M = 12 \]

### Final Results
- The resolving limit of the telescope is \(6.1 \times 10^{-6} \, \text{radians}\).
- The magnifying power of the telescope is \(12\).