Question

A monochromatic light of wavelength 6000 \AA \ coming from a star is detected in a 100 inch telescope. The limit of resolution of the telescope is approximately

• A. $3.4 \times 10^{-7}$ rad
• B. $6.7 \times 10^{-7}$ rad
• C. $2.9 \times 10^{-7}$ rad
• D. $1.54 \times 10^{-7}$ rad

Hint:

Rayleigh criterion for angular resolution: $\theta_{min} \approx 1.22 \frac{\lambda}{D}$.
Ensure all length units ($\lambda$ and $D$) are consistent (e.g., meters).
$1 \text{ \AA} = 10^{-10} \text{ m}$.
$1 \text{ inch} = 2.54 \text{ cm} = 0.0254 \text{ m}$.
The result for $\theta_{min}$ will be in radians.

Answer 1

The Correct Option is C

The limit of resolution (angular resolution) of a telescope with a circular aperture is given by the Rayleigh criterion: $\theta_{min} \approx 1.22 \frac{\lambda}{D}$ where $\theta_{min}$ is the minimum resolvable angle (in radians), $\lambda$ is the wavelength of light, and $D$ is the diameter of the telescope's objective lens or mirror. Given values: Wavelength $\lambda = 6000 \text{ \AA} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$. Diameter of the telescope $D = 100 \text{ inches}$. We need to convert inches to meters. $1 \text{ inch} = 2.54 \text{ cm} = 0.0254 \text{ m}$. So, $D = 100 \times 0.0254 \text{ m} = 2.54 \text{ m}$. Now, substitute these values into the formula for $\theta_{min}$: $\theta_{min} \approx 1.22 \times \frac{6 \times 10^{-7} \text{ m}}{2.54 \text{ m}}$. $\theta_{min} \approx 1.22 \times \frac{6}{2.54} \times 10^{-7} \text{ rad}$. $\frac{6}{2.54} \approx 2.3622$. $\theta_{min} \approx 1.22 \times 2.3622 \times 10^{-7} \text{ rad}$. $1.22 \times 2.3622 \approx 2.881884$. So, $\theta_{min} \approx 2.88 \times 10^{-7} \text{ rad}$. This value is closest to option (c) $2.9 \times 10^{-7}$ rad. \[ \boxed{2.9 \times 10^{-7} \text{ rad}} \]