A 7 by 50 binocular has objectives of diameter mm and angular magnification

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Geometrical optics and imaging · Geometrical optics and imaging

Optics and Photonics, Ch Radiometry and photometry

Problem

A 7 by 50 binocular has objectives of diameter D=50D = 50 mm and angular magnification ∣MA∣=7|M_A| = 7, so its exit pupil diameter is D/∣MA∣D/|M_A| (Chapter 2). Its true field of view has half-angle 0.50∘0.50^\circ; the apparent field half-angle at the eye is therefore 77 times as large. The instrument is used in daylight, when the observer's eye pupil is 2.02.0 mm across. Take the instrument as lossless and everything as being in air, n=1.00n = 1.00, and use the narrow-cone form Ω=πu2\Omega = \pi u^2 with uu in radians. ((a)) Find the exit pupil diameter. ((b)) Find the etendue G=n2AΩG = n^2 A\Omega of the beam leaving the exit pupil, using the exit pupil area and the apparent field's solid angle. ((c)) Find the etendue the eye can accept, using its own pupil area and the same apparent field. ((d)) Hence find the fraction of the flux gathered by the objectives that reaches the retina. ((e)) What objective diameter would waste no etendue at this magnification and eye pupil?

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