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Diffraction of Light

Problem 5.150

Find the minimum magnification of a microscope, whose objective's numerical aperture is sin⁡α=0.24,\sin \alpha=0.24, at which the resolving power of the objective is totally employed if the diameter of the eye's pupil is d0=4.0 mmd_{0}=4.0 \mathrm{~mm}

Reveal Answer
5.150. Suppose dmin⁡d_{\min } is the minimum separation resolved by the microscope's objective, Δψ\Delta \psi is the angle subtended by the eye at the object over the distance of the best visibility l0(25 cm),l_{0}(25 \mathrm{~cm}), and Δψ′\Delta \psi^{\prime} is the minimum angular separation resolved by the eye (Δψ′=\left(\Delta \psi^{\prime}=\right. =1.22λ/d0)\left.=1.22 \lambda / d_{0}\right). Then the sought magnification of the microscope is Γmin⁡=Δψ′/Δψ=2(l0/d0)sin⁡α=30\Gamma_{\min }=\Delta \psi^{\prime} / \Delta \psi=2\left(l_{0} / d_{0}\right) \sin \alpha=30
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