Question

Converging lens of telescope \(5\) cm in diameter has focal length \(25\) cm. In the focal plane of the lens the distance between the centres of the Fraunhofer diffraction pattern is (wavelength of light used = \(5000\) Å, Rayleigh's criterion is satisfied)

• A. \(25000\) Å
• B. \(28000\) Å
• C. \(30500\) Å
• D. \(32000\) Å

Hint:

Rayleigh criterion: angular limit 1.22 lambda / D; linear separation = f times angle.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Two point sources are just resolved when the angular separation equals \(\theta=\dfrac{1.22\lambda}{D}\), where \(D\) is the lens diameter.

Step 2: Compute the angle:
\(\lambda=5000\) \(\text{\AA}\) \(=5\times10^{-7}\) m, \(D=5\) cm \(=0.05\) m. \(\theta=\dfrac{1.22\times5\times10^{-7}}{0.05}=1.22\times10^{-5}\) rad.

Step 3: Find the linear separation at the focal plane:
\(y=f\theta=0.25\times1.22\times10^{-5}=3.05\times10^{-6}\) m \(=30500\) \(\text{\AA}\). Option C.

Step 4: Why the other options are wrong.
Without the 1.22 factor the answer would be 25000 \(\text{\AA}\) (option A), which is the wrong criterion for a circular aperture.

Final Answer:
The separation is 30500 angstrom. \[ \boxed{\text{(C) }30500\ \text{\AA}} \]