Collecting Power of Telescopes (AQA A Level Physics): Revision Note

Collecting Power

• Telescopes are designed to gather as much light as possible

  • The more light energy a telescope can gather, the brighter the images it will be able to produce

  • This can be measured by a telescope's collecting power

• The collecting power of a telescope is defined as:

A measure of the amount of light energy it collects per second

• This is equivalent to the power per unit area, or intensity of the incident radiation collected

Differences in Collecting Power

The telescope with the larger aperture is able to produce a brighter image

• The collecting power of a telescope is directly proportional to the square of the diameter of its objective

$\mathrm{collecting}\mathrm{power}\propto D^2$

• This is because: 

  • Intensity is proportional to surface area 

  • The surface area of a circular object of diameter D is equal to $\frac{\mathrm{\pi }{D}^{\mathit{2}}}{4}$

• This means that objects at greater distances can also be seen

  • The intensity of light from a point source decreases inversely with the square of the distance according to the inverse square law

Large Diameter Telescopes

• Larger aperture diameter telescopes are advantageous for two main reasons:

  • They have a greater collecting power, so images are brighter

  • They have a greater resolving power, so images are clearer

• The collecting power of two telescopes can be calculated using the ratio

$\frac{\mathrm{collecting}\mathrm{power}\mathrm{of}\mathrm{telescope}1}{\mathrm{collecting}\mathrm{power}\mathrm{of}\mathrm{telescope}2}={\left(\frac{D_1}{D_2}\right)}^2$

• The resolving power of two telescopes operating at the same wavelength can be calculated using the ratio

$\frac{\mathrm{resolving}\mathrm{power}\mathrm{of}\mathrm{telescope}1}{\mathrm{resolving}\mathrm{power}\mathrm{of}\mathrm{telescope}2}=\frac{{\theta }_1}{{\theta }_2}=\frac{{\displaystyle \frac{\lambda }{D_1}}}{{\displaystyle \frac{\lambda }{D_2}}}=\frac{D_2}{D_1}$