Question:
A monochromatic light of wavelength \(6000\ \text{\AA}\) coming from a star is detected in a \(100\)-inch telescope. The limit of resolution of the telescope is approximately
A monochromatic light of wavelength \(6000\ \text{\AA}\) coming from a star is detected in a \(100\)-inch telescope. The limit of resolution of the telescope is approximately
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For a telescope,
\[
\theta=\frac{1.22\lambda}{D}.
\]
A larger objective diameter \(D\) gives a smaller value of \(\theta\), resulting in better resolving power.
Updated On: Jun 26, 2026
- \(3.4\times10^{-7}\ \text{rad}\)
- \(6.7\times10^{-7}\ \text{rad}\)
- \(2.9\times10^{-7}\ \text{rad}\)
- \(1.54\times10^{-7}\ \text{rad}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use Rayleigh's criterion for a telescope.
The angular limit of resolution of a telescope is \[ \theta=\frac{1.22\lambda}{D}, \] where \[ \lambda \] is the wavelength of light and \[ D \] is the diameter of the objective lens or mirror.
Step 2: Convert the given quantities into SI units.
Given wavelength, \[ \lambda=6000\ \text{\AA}. \] Since \[ 1\ \text{\AA}=10^{-10}\ \text{m}, \] \[ \lambda=6000\times10^{-10}. \] \[ \lambda=6\times10^{-7}\ \text{m}. \] Diameter of telescope, \[ D=100\ \text{inch}. \] Since \[ 1\ \text{inch}=2.54\times10^{-2}\ \text{m}, \] \[ D=100\times2.54\times10^{-2}. \] \[ D=2.54\ \text{m}. \]
Step 3: Calculate the limit of resolution.
\[ \theta=\frac{1.22(6\times10^{-7})}{2.54}. \] \[ \theta=\frac{7.32\times10^{-7}}{2.54}. \] \[ \theta\approx2.88\times10^{-7}\ \text{rad}. \] \[ \theta\approx2.9\times10^{-7}\ \text{rad}. \]
Step 4: Final conclusion.
Hence, the limit of resolution of the telescope is \[ \boxed{2.9\times10^{-7}\ \text{rad}} \] Therefore, the correct option is \[ \boxed{(3)} \]
The angular limit of resolution of a telescope is \[ \theta=\frac{1.22\lambda}{D}, \] where \[ \lambda \] is the wavelength of light and \[ D \] is the diameter of the objective lens or mirror.
Step 2: Convert the given quantities into SI units.
Given wavelength, \[ \lambda=6000\ \text{\AA}. \] Since \[ 1\ \text{\AA}=10^{-10}\ \text{m}, \] \[ \lambda=6000\times10^{-10}. \] \[ \lambda=6\times10^{-7}\ \text{m}. \] Diameter of telescope, \[ D=100\ \text{inch}. \] Since \[ 1\ \text{inch}=2.54\times10^{-2}\ \text{m}, \] \[ D=100\times2.54\times10^{-2}. \] \[ D=2.54\ \text{m}. \]
Step 3: Calculate the limit of resolution.
\[ \theta=\frac{1.22(6\times10^{-7})}{2.54}. \] \[ \theta=\frac{7.32\times10^{-7}}{2.54}. \] \[ \theta\approx2.88\times10^{-7}\ \text{rad}. \] \[ \theta\approx2.9\times10^{-7}\ \text{rad}. \]
Step 4: Final conclusion.
Hence, the limit of resolution of the telescope is \[ \boxed{2.9\times10^{-7}\ \text{rad}} \] Therefore, the correct option is \[ \boxed{(3)} \]
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