An astronomical telescope has its two lenses spaced 81.1 cm apart. If the objective lens has a focal length of 78.0 cm , what is the magnifi

Question

An astronomical telescope has its two lenses spaced 81.1 cm apart. If the objective lens has a focal length of 78.0 cm , what is the magnification of this telescope? Assume a relaxed eye. Follow the sign conventions.

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Eirian 1 month 2021-08-12T08:03:37+00:00 1 Answers 1 views 0

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    2021-08-12T08:04:52+00:00

    Answer:

    The magnification is  m = -26.16

    Explanation:

    From the question we are told that

            The distance between the lenses is  L =81.1 cm

             The focal length of the objective lens f_o = 78.0 \ cm

    Generally the  length of the telescope that satisfies the condition for a relaxed eye can be mathematically represented as

                  L = f_0 + f_e

    Where f_e is focal length of the eyepiece

          Making f_e the subject the of the formula above

                    f_e = L - f_o

    Now substituting value

                   f_e = 81.1 - 78.0

                       =3.1\ cm

    Generally magnification can be mathematically represented as

                                m = - \frac{f_o}{f_e}

    This negative sign is a reminder that all real image are always inverted

                                      =-  \frac{81.1}{3.1}

                                      = - 26.16  

                   

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )