MIT2_71S09_gsol2

MIT2_71S09_gsol2 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Solutions to Problem Set #2 Due Wednesday, Feb. 25, 2009 Problem 1: Wiper speed control Figure 1 shows an example of an optical system designed to detect the amount of water present on the windshield of a car to adjust the wiper speed. As shown in this at one end (bottom of the windshield) to a detector located in the opposite end. The light su/ers total-internal re±ection (TIR) at the glass-air interface. However, when rain drops are present, some of the light will su/er frustrated TIR escaping outside the waveguide. Since we know the power of the light source, a given drop in power can be correlated to the amount of water present and used to adjust the wiper speed. To quantify the feasibility of this design we start by computing the critical angles of the glass-air and glass-water interfaces, air c g a = csin n ar = n glass ± 41 : 8 ; c g w = arcsin n water 62 n ± = : 5 : glass The incidence angle of a given ray propagating inside this waveguide is restricted to the following cases: 1. For c g a : The light will su/er frustrated TIR and escape out of the waveguide regardless of whether the interface is glass-air or glass-water. 2. For c g a c g w : The light will su/er TIR at the glass-air interface and frustrated TIR at the glass-water interface. 3. For c g w : The light will su/er TIR at both interfaces. Problem 2: Telescope with in±nite conjugates a) The geometry of this problem is shown in Figure 2. By inspection, we see that f 1 at its back focal plane. This point now acts as a point source object for the second lens (L2) and since it is located at its front focal plane (provided that d = f 1 + f 2 ), the light beam gets collimated and thus exits the optical system as a parallel ray bundle (plane wave). b) To answer this question we use the matrix formulation. The output angle and height are related to the input angle and height according to, 1
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Figure 1: Wiper speed control problem. 2
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2 x 2 ± = 1 1 1 f 2 0 1 1 0 d 1 1 f 1 0 1 1 x 1 ± (1) = 1 d d f 2 1 f 1 f 2 1 f 1 f 2 d 1 d : f ± 1 1 1 x ± So the optical power of the composite optical system is, d P = 1 f 1 f 2 1 f 1 ( f = f 2 1 + f 2 ) d ; (2) f 1 f 2 and the e/ective focal length is, 1 EFL = f 1 f 2 = P : (3) ( f 1 + f 2 ) d Note that for the case of d = f 1 + f 2 , equations 2 and 3 become: P
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MIT2_71S09_gsol2 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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