Answer:
[tex]d_{eye\ piece} =1.93\ mm[/tex]
Explanation: Â Â Â Â Â Â Â Â Â Â
angular magnification (m_θ) = 43            Â
diameter of the objective = 83 mm       Â
                     = 0.083 m
The minimum diameter of eyepiece            Â
[tex]d_{eye\ piece} = \dfrac{d_{object}}{m_\theta}[/tex]
[tex]d_{eye\ piece} = \dfrac{83}{43}[/tex] Â Â Â Â Â Â Â Â Â Â Â Â
[tex]d_{eye\ piece} =1.93\ mm[/tex] Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
Hence, The minimum diameter of the eyepiece required = 1.93 mm
