Solutions To Spherical Mirror - Reflecting Telescope Problems (Practical Astronomy Focus)

 The Problems:

1) A small object lies 4 cm to the left of the vertex of a concave mirror who radius of curvature is 12 cm. Find the magnification of the image.  Include a basic sketch to show the relation of the image to the object position relative to the mirror.

2) What would be the size of the image of the Moon produced by the 14-inch (356 mm) Harry Bayley Observatory Cassegrain telescope? (Focal length = 154 inches)  What would be the radius of curvature R for the telescope mirror?

3) Repeat the above computations for a 6-inch Newtonian reflecting telescope with a focal length of 48 inches.

Solutions:

(1) We apply the mirror equation: 1/s  + 1/s’ = 2 /R

Where: s = 4 cm, which is positive (to left of mirror). Also, we have R = 12 cm. Then we need to find s’ the image distance:

1/s’ =  2/ R – 1/s = 2/12 – ¼ = 1/6 – ¼ 

For which s’ = -12 cm

The negative sign denotes a virtual image which lies to the right of the mirror and this situation can be depicted as shown below:

The magnification of the image is just: M = s’/s = h’/ h

So: M =  ï s’/s ï  = ï -12cm /4 cm ï    = 4

(2) Using:    s = 0.01744f

In normal application this applies to an image corresponding to  1 ^o  in the sky.

For the Harry Bayley Observatory Schmidt-Cassegrain:  f= 154 in. we get s = 0.01744 (154 in.) =  2. 7 inches per degree.

But the Moon subtends  1/2 ^o  so that the resulting image scale for the HBO  telescope would be 1/2 what it is for one degree in the sky, or:

½ (2.7/ deg) = 1.35 in/ deg

(3) For the cited Newtonian reflector, f =  48 in. 

For which we get s = 0.01744 (48 in.) =  0.84  inches per degree.

But the Moon subtends  1/2 ^o  so that the resulting image scale for the HBO  telescope would be 1/2 what it is for one degree in the sky, or:

½ (0.83/ deg) = 0.42 in/ deg