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

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?

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Diễm Kiều 3 years 2021-08-16T03:44:48+00:00 1 Answers 16 views 0

Answers ( )

  1. Neala
    0
    2021-08-16T03:46:20+00:00
    Reply

    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

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