Question

A telescope using light having wavelength $5000\mathring{A}$ and using lenses of focal lengths $2.5\text{cm}$ and $30\text{cm}$. If the diameter of the aperture of the objective is $10\text{cm}$, then the resolving power of the telescope is-

• A. $2.55\times {10}^5{\text{rad}}^{-1}$
• B. $1.64\times {10}^5{\text{rad}}^{-1}$
• C. $3.68\times {10}^6{\text{rad}}^{-1}$
• D. $9.12\times {10}^6{\text{rad}}^{-1}$

Answer 1

The correct option is B $1.64\times {10}^5{\text{rad}}^{-1}$

Given,

$\lambda =5000\mathring{A}=5000\times {10}^{-10}\text{m};a=10\text{cm}{f}_{\circ }=30\text{cm};f_e=2.5\text{cm}$

The resolving power of a telescope is,

$R.P.=\frac{a}{1.22\lambda }$

Where,

$a=\text{Diameter (aperture) of the objective lens and}\lambda =\text{Wavelength of the light}$

$\Rightarrow R.P.=\frac{10\times {10}^{-2}}{1.22\times 5\times {10}^{-7}}$

$R.P.\approx 1.64\times {10}^5{\text{rad}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.

Why this question ? Key concept :Resolving power of a telescope is, $R.P.=\frac{a}{1.22\lambda }$ where, $a=\text{Diameter (aperture) of the objective lens and}\lambda =\text{Wavelength of the light}$