Question

Resolving power of a telescope is \(5\times10^{-7}\,\text{rad}\). If wavelength of incident light is \(500\,\text{nm}\), find the diameter of the aperture of the telescope.

• A. \(1.22\,\text{m}\)
• B. \(4.66\,\text{m}\)
• C. \(2.33\,\text{m}\)
• D. \(0.56\,\text{m}\)

Answer 1

The Correct Option is A

Concept:
Resolving power of a telescope is given by Rayleigh’s criterion: \[ \theta = \frac{1.22\lambda}{D} \] where \(\theta\) = angular resolution, \(\lambda\) = wavelength of light, \(D\) = diameter of the objective lens. Step 1: Substitute the given values. \[ \theta = 5\times10^{-7}\,\text{rad} \] \[ \lambda = 500\,\text{nm} = 500\times10^{-9}\,\text{m} \] \[ 5\times10^{-7} = \frac{1.22\times500\times10^{-9}}{D} \] Step 2: Solve for \(D\). \[ D = \frac{1.22\times500\times10^{-9}}{5\times10^{-7}} \] \[ D = 1.22\,\text{m} \]