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Calculate the smallest angular separation resolved by the human eye, given : aperature =2.5mm and effectived lambda= 5500 dotA. If a scale with mm markings is viewed by the unaided eye, deduce the largest distance to which the markings will be visible. |
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Answer» Solution :Here, `lambda 5500 DOTA= 5500xx 10^(-10)m` `a= 2.5mm= 2.5xx 10^(-3)m` `triangle THETA = 1.22(lambda)/(a)= 1.22((5500xx10^(-10)m)/(2.5xx 10^(-3)m))= 2.7xx 10^(-7)rad""(i)` angle subtended by 1mm markings from a distance D (in metres) `i.e., theta= (d)/(D)= (1xx 10^(-3)m)/(D)= (1)/(D)xx 10^(-3)"""........"(II)` From eqns. (i) and (ii), `(1)/(D)xx10^(-3)= 2.7xx 10^(-4)"(or) "D= (10^(-3))/(2.7xx 10^(-4))= 3.7m`. |
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