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202   Chapter Nine

        in 10 in it drops to a height of 0.5   10   0.0227272   0.272727 in. The “real”
        field is thus  0.0272727 and the “apparent” field is  0.272727.
          For (approximately) 50 percent vignetting, one would use the principal ray
        through the center of the aperture stop.

         4 A 4-in focal length f/4 lens is used to project an image at a magnification
        of four times (m   4). What is the numerical aperture (NA) in object space,
        and in image space?
        ANSWER:  The clear aperture of the lens is 4 in/4   1 in. Using Eq. 2.6, and
        solving for x and x′ we get
                x   f/m  and      x′   fm which gives us
                x   4/( 4)   1 in    and      x′   4( 4)   16 in
        and     s   x   f   5 in     and      s′   x′   f   16  4   20 in
                                            1
        If we ignore trigonometry and use NA   
 CA/s, for object space the NA is
                                             2
        0.5/5   0.1 and in image space NA   0.5/20   0.025.
         5 An optical system composed of two thin elements forms an image of an
        object at infinity. The front lens has a 16-in focal length, the rear lens has an
        8-in focal length and the lenses are separated by 8 in. If the exit pupil is located
        at the rear lens and there is no vignetting, what is the illumination at an
        image point 3 in from the axis, relative to the illumination on the axis?
        ANSWER:  To locate the image plane we can use (among many other possibili-
        ties) Eq. 4.6:
            B   f (f   d)/(f   f   d)   8(16   8)/(16   8   8)   64/16   4 in.
                  a
                b
                             b
                         a
        This is the stop to image distance. For a point 3 in from the axis the ray slope
        is arctan (3/4)   36.863898  and its cosine is 0.8. The cos is 0.4096 and the
                                                         4
        relative illumination is 41 percent.
         6 A 6-in diameter f/5 paraboloid is part of an infrared tracker which can tol-
        erate a blur (due to defocusing) of 0.1 milliradians. (a) What tolerance must
        be maintained on the position of the reticle with respect to the focal point?
        (b) What is the tolerance if the system has a speed of f/2?
        ANSWER: (a) The focal length of the paraboloid is 6   5   30 in and a 0.1 milli-
        radian blur has a diameter of 30   0.0001   0.003 in. At a speed of f/5 a blur
        of 0.003 in will be generated by a defocus of 0.003   5   0.015 in.
        (b) At a speed of f/2 the focal length is 12 in and the blur is 12   0.0001
        0.0012 in. At f/2 the defocus is 2   0.0012   0.0024 in.

         7 The hyperfocal distance of a 10-in focal length, f/10 lens is 100-in. (a) What
        is the diameter of the “acceptable” blur spot? (b) what is the closest distance
        at which an object is “acceptably” in focus? (c) Show that the answer to (b) is
        always one half the hyperfocal distance.
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