Question

A small Keplerian telescope has an objective with a 1.33 m focal length. Its eyepiece is a 2.82 cm focal length lens. It is used to look at a 25000 km diameter sunspot on the sun, a distance 1.5*108 km from Earth.
Required:
What angle is subtended by the sunspot's telescopic image in degree?

Answer 1

Answer:

The angle is  $\phi = 0.45 0 ^o$

Explanation:

From the question we are told that  

     The objective  focal length   $f = 1.33 \ m$

     The  eyepiece focal length is  $f_o = 2.82 \ cm = 0.0282 \ m$

      The diameter of the sunlight is  $d = 25000km = 2.5 *10^{7} \ m$

       The distance of the sun from from the earth is  $D = 1.5 *10^8 km = 1.5 *10^{11} \ m$

  Generally the magnification of the object is mathematically evaluated as

        $m = -\frac{f_o }{f_e }$

The negative  sign is because the lens of the telescope is  diverging light

 substituting values  

      $m = -\frac{1.33 }{0.0282 }$

      $m = - 47.16 3$

Now we can obtain the angle made by the object (sunlight ) with respect to the telescope  as follows  

        $tan \theta = \frac{d}{D}$

substituting values

      $tan \theta = \frac{2.5 *10^{7}}{1.5*10^{11}}$

      $tan \theta = 0.0001667$

      $\theta= tan^{-1}[0.0001667]$

     $\theta= 0.00955^o$

The  magnification can  also be mathematically represented as

      $m = \frac{\phi }{\theta }$

Where $\phi$ is the angle the image made with telescope

Since the negative sign indicate direction of light movement we will remove it from the calculation below

      =>   $47.163 = \frac{\phi}{0.00955}$

     =>    $\phi = 0.45 0 ^o$