Question

A telescope with objective diameter \(R\) is used to observe a distant star emitting light of wavelength 500 nm, at a resolution of \(5 \times 10^{-7}\) radian. The value of \(R\) is ________ cm.

• A. 61
• B. 122
• C. 244
• D. 305

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The resolving power of a telescope is determined by the Rayleigh criterion. The limit of resolution (minimum angular separation \(\Delta \theta\)) is the smallest angle between two distant objects that can just be seen as separate.

Step 2: Key Formula or Approach:
1. Rayleigh Criterion: \(\Delta \theta = \frac{1.22 \lambda}{R}\) 2. Here, \(\lambda\) is the wavelength, \(R\) is the diameter of the objective lens, and \(\Delta \theta\) is the resolution in radians.

Step 3: Detailed Explanation:
1. Given: \(\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}\), \(\Delta \theta = 5 \times 10^{-7} \text{ rad}\). 2. Rearrange the formula to solve for \(R\): \[ R = \frac{1.22 \lambda}{\Delta \theta} \] 3. Substitute the values: \[ R = \frac{1.22 \times 500 \times 10^{-9}}{5 \times 10^{-7}} \] 4. Simplify the calculation: \[ R = \frac{1.22 \times 5 \times 10^{-7}}{5 \times 10^{-7}} = 1.22 \text{ m} \] 5. Convert meters to centimeters: \[ R = 1.22 \times 100 = 122 \text{ cm} \]

Step 4: Final Answer:
The value of the objective diameter \(R\) is 122 cm.