Answer:
The angle is [tex]\phi = 0.45 0 ^o[/tex]
Explanation:
From the question we are told that
The objective focal length [tex]f = 1.33 \ m[/tex]
The eyepiece focal length is [tex]f_o = 2.82 \ cm = 0.0282 \ m[/tex]
The diameter of the sunlight is [tex]d = 25000km = 2.5 *10^{7} \ m[/tex]
The distance of the sun from from the earth is [tex]D = 1.5 *10^8 km = 1.5 *10^{11} \ m[/tex]
Generally the magnification of the object is mathematically evaluated as
[tex]m = -\frac{f_o }{f_e }[/tex]
The negative sign is because the lens of the telescope is diverging light
substituting values
[tex]m = -\frac{1.33 }{0.0282 }[/tex]
[tex]m = - 47.16 3[/tex]
Now we can obtain the angle made by the object (sunlight ) with respect to the telescope as follows
[tex]tan \theta = \frac{d}{D}[/tex]
substituting values
[tex]tan \theta = \frac{2.5 *10^{7}}{1.5*10^{11}}[/tex]
[tex]tan \theta = 0.0001667[/tex]
[tex]\theta= tan^{-1}[0.0001667][/tex]
[tex]\theta= 0.00955^o[/tex]
The magnification can also be mathematically represented as
[tex]m = \frac{\phi }{\theta }[/tex]
Where [tex]\phi[/tex] 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
=> [tex]47.163 = \frac{\phi}{0.00955}[/tex]
=> [tex]\phi = 0.45 0 ^o[/tex]
