All Problems Diffraction of Light
Problem 5.150
Find the minimum magnification of a microscope, whose objective's numerical aperture is \(\sin \alpha=0.24,\) at which the resolving power of the objective is totally employed if the diameter of the eye's pupil is \(d_{0}=4.0 \mathrm{~mm}\)
5.150. Suppose \(d_{\min }\) is the minimum separation resolved by the microscope's objective, \(\Delta \psi\) is the angle subtended by the eye at the object over the distance of the best visibility \(l_{0}(25 \mathrm{~cm}),\) and \(\Delta \psi^{\prime}\) is the minimum angular separation resolved by the eye \(\left(\Delta \psi^{\prime}=\right.\) \(\left.=1.22 \lambda / d_{0}\right)\). Then the sought magnification of the microscope is \(\Gamma_{\min }=\Delta \psi^{\prime} / \Delta \psi=2\left(l_{0} / d_{0}\right) \sin \alpha=30\)