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