Page 220 - Modern Optical Engineering The Design of Optical Systems
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Stops, Apertures, Pupils and Diffraction  203

        ANSWER:  (a) The lens diameter is 10 in/10   1 in and per Eq. 2.3 the image of
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        an object at 100 in from the lens is x′   f /x   100/( 90)   1.111 . . . in
        from the second focal point. An object at infinity is imaged at the focal point,
        and at f/10 the blur is 1.111/10   0.1111 in.
        (b) The closest “acceptable” distance object will be imaged   beyond the image for
        the hyperfocal distance. By similar triangles:  /0.1111   (10   1.111   )/0.5 in
        and    1.3888, or at 2.5 in beyond the focal point. The corresponding object
        distance is x   100/( 2.5)   40 in and the object distance from the lens is
        40   10   50 in, or one-half the hyperfocal distance.
        (c) This exercise is left to the reader.

         8 Compare the image illumination produced by an f/8 lens at a point 45°
        from the axis with that from an f/16 lens 30° off axis.
        ANSWER:  Illumination varies inversely with the square of the f-number and
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        directly as the fourth power of the cosine of the angle of obliquity, or cos /(f/#) .
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        For the f/8 lens this is cos 45 /64   0.25/64   0.003906 and for the f/6 lens it
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        is cos 30 /256    0.5625/256    0.002197. Thus the  f/8 lens illumination is
        0.003906/0.002197   1.777 . . . , or 78 percent higher than the f/16 lens.
         9 An optical system is required to image a distant point source as a spot
        0.01 mm in diameter. Assuming that all the useful energy in the diffraction
        pattern will be within the first dark ring, for a wavelength of 550 nm what rel-
        ative aperture and numerical aperture must the system have?
        ANSWER:  Per Eq. 9.20 the radius of the first dark ring of the diffraction pat-
        tern is equal to 1.22	 (f/#)    0.61	/NA. So the NA must equal 2    0.61
        .00055/0.01   0.0671 and the f/# is f/7.45.
        10 A pinhole camera has no lens but uses a very small hole some distance
        from the film to form its image. If we assume that light travels in straight
        lines, then the image of a distant point source will be a blur whose diameter
        is the same as the pinhole. However, diffraction will spread the light into an
        Airy disk. Thus the larger the hole the bigger the geometrical blur, but the
        smaller the diffraction pattern. Assume that the sharpest picture will be pro-
        duced when the geometrical blur is the same size as the central bright spot of
        the diffraction pattern. What size hole should be used when the film is 100 mm
        from hole? (Hint: Equate the hole diameter to the diameter of the first dark
        ring of the Airy pattern, as given by Eq. 9.20.)
        ANSWER:  Diameter   2   1.22	 (f/#) where f/#   100 mm/Diameter.
        Thus D   2   1.22   0.00055   100/D
        and D 5 20.1342  5  0.366 mm

        11 What is the resolution limit (at the object) for a microscope objective
        whose acceptance cone has a numerical aperture of (a) 0.25, (b) 0.80, (c) 1.2 at
        a wavelength of 550 nm?
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