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 ( )

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    2021-08-16T03:46:20+00:00

    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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