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Photometry and Geometrical Optics

Problem 5.49

A microscope has a numerical aperture \(\sin \alpha=0.12,\) where \(\alpha\) is the aperture angle subtended by the entrance pupil of the microscope. Assuming the diameter of an eye's pupil to be equal to \(d_{0}\) \(=4.0 \mathrm{~mm},\) determine the microscope magnification at which (a) the diameter of the beam of light coming from the microscope is equal to the diameter of the eye's pupil; (b) the illuminance of the image on the retina is independent of magnification (consider the case when the beam of light passing through the system "microscope-eye" is bounded by the mounting of the objective).

Reveal Answer
5.49. (a) Γ=2αl0/d0=15, where l0 is the distance of the best  vision (25 cm);(b)Γ⩽2αl0/d0\begin{aligned} &5.49 . \text { (a) } \Gamma=2 \alpha l_{0} / d_{0}=15, \text { where } l_{0} \text { is the distance of the best }\\ &\text { vision }(25 \mathrm{~cm}) ;(\mathrm{b}) \Gamma \leqslant 2 \alpha l_{0} / d_{0} \end{aligned}
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