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Step-by-Step Solution
Step 1: Identify the relevant formula
The minimum angular separation $ \Delta \theta $ for two point objects (stars) to be just resolved by a telescope is given by:
$ \Delta \theta = \frac{1.22 \,\lambda}{D} $
where:
$ \lambda $ is the wavelength of light used.
$ D $ is the diameter of the telescope’s objective.
The factor 1.22 comes from the diffraction pattern (Airy disk) for a circular aperture.
Also, we use $ \Delta \theta \approx \frac{l}{R} $ when $ l $ is the linear separation between the two stars and $ R $ is their distance from the telescope.
Step 2: Rearrange to find the linear separation
From $ \Delta \theta = \frac{l}{R} $ and $ \Delta \theta = \frac{1.22\,\lambda}{D} $, we get:
$ l = \frac{1.22 \, \lambda \, R}{D} \,.$
Step 3: Substitute the given values
Wavelength, $ \lambda = 6 \times 10^{-7} \, \text{m} $
Diameter of telescope, $ D = 30 \, \text{cm} = 30 \times 10^{-2} \, \text{m} $
Distance of the stars, $ R = 10 \, \text{light years} = 10 \times 9.46 \times 10^{15} \, \text{m} $
Numerical factor, 1.22
Therefore,
$
l
=
\frac{1.22 \times \left(6 \times 10^{-7}\right) \times \left(10 \times 9.46 \times 10^{15}\right)}{30 \times 10^{-2}}
\,.
$
Step 4: Perform the calculation
First, simplify the numerator:
$
\text{Numerator}
=
1.22 \times 6 \times 10^{-7} \times 10 \times 9.46 \times 10^{15}
=
1.22 \times 6 \times 9.46 \times 10 \times 10^{-7} \times 10^{15}
=
(1.22 \times 6 \times 9.46 \times 10) \times 10^{(15 - 7)}
\,.
$
$ =
(1.22 \times 6 \times 9.46 \times 10) \times 10^8
\,.
$
Calculate the factor separately (approximate):
$ 1.22 \times 6 \times 9.46 \times 10
\approx
1.22 \times 6 \times 9.46 \times 10
\approx
(1.22 \times 9.46) \times (6 \times 10)
\approx
11.5412 \times 60
\approx
692.472
\,.
$
So the numerator is about $ 692.472 \times 10^8 = 6.92472 \times 10^{10} \, \text{m}. $
The denominator is $ 30 \times 10^{-2} = 0.30 \, \text{m}. $
Hence,
$
l
=
\frac{6.92472 \times 10^{10} \, \text{m}}{0.30}
=
2.30824 \times 10^{11} \, \text{m}
\,.
$
Step 5: Convert meters to kilometers
$ 1 \, \text{km} = 10^3 \, \text{m}. $ Hence,
$
l
=
2.30824 \times 10^{11} \, \text{m}
\times
\frac{1 \, \text{km}}{10^3 \, \text{m}}
=
2.30824 \times 10^{8} \, \text{km}
\,.
$
The result is of the order of $ 10^8 \, \text{km}. $
Final Answer
The minimum distance between the two stars, to be just resolved by the telescope, is of the order of $ 10^8 \, \text{km}. $
