p37_064

p37_064 - sin tan to relate these directly, or we could be...

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64. Consider two light rays crossing each other at the middle of the lens (see Fig. 37-42(c)). The rays come from opposite sides of the circular dot of diameter D , a distance L from the eyes, so we are using the same notation found in Sample Problem 37-6 (which is in the textbook supplement). Those two rays reach the retina a distance L 0 behind the lens, striking two points there which are a distance D 0 apart. Therefore, D L = D 0 L 0 where D =2mmand L 0 = 20 mm. If we estimate L 450 mm, we ±nd D 0 0 . 09 mm. Turning our attention to Fig. 37-42(d), we see θ =tan 1 µ 1 2 D 0 x which we wish to set equal to the angle in Eq. 37-12. We could use the small angle approximation
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Unformatted text preview: sin tan to relate these directly, or we could be exact as we show below: If tan = b a , then sin = b a 2 + b 2 . Therefore, this exact use of Eq. 37-12 leads to 1 . 22 d = sin = 1 2 D p x 2 + ( D / 2) 2 where = 550 10 6 mm and 1 mm x 15 mm. Using the value of D found above, this leads to a range of d values: 0 . 015 mm d . 23 mm....
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