Question

The least resolvable angle by a telescope using an objective of aperture $5m$ is nearly $(\lambda =4000\AA )$

• A. $\frac{1}{50°}$
• B. $(1/50)$minutes
• C. $(1/50)$sec
• D. $(1/500)$sec

Answer 1

The correct option is C

$(1/50)$sec

Step 1: Given Data

The objective aperture is $d$ is $5m$

Wavelength is $\lambda =4000\AA$

Step 2: Formula used,

${\theta }_{min}=\frac{1.22\lambda }{d}$

Where $\lambda$ is the wavelength

$d$ is the objective aperture

Step 3: Calculation

⇒ ${\theta }_{min}=\frac{1.22(4000\times {10}^{-10})}{5}$

⇒ ${\theta }_{min}=800x1.22x{10}^{-10}rad$

now we have to convert radians into degrees

$1rad=\frac{180}{\mathrm{\pi }}deg$

⇒ ${\theta }_{min}=800x1.22x{10}^{-10}\times \frac{180}{\mathrm{\pi }}deg$

$1°=60min=360sec$

⇒ ${\theta }_{min}=800x1.22x{10}^{-10}\times \frac{180}{\mathrm{\pi }}\times 360sec$

⇒ ${\theta }_{min}=\frac{1}{50}sec$

Hence, the correct answer is option C