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Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil diameter 3mm. Approximately, what is the maximum distance at which these dits can be resolved by the eye? [Take wavelelngth of light =500nm]A. `1 m`B. `5 m`C. `3 m`D. `6 m` |
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Answer» Correct Answer - B Limit of resolution eye `= theta = (1.22 lambda)/(D)` `= (1.22 xx 5 xx 10^(-7))/(3 xx 10^(-3)) = 2.03 xx 10^(-4) rad` If `x` is the minimum distance at width dots are just resolved, then `theta = (1 mm)/(x) = (10^(-3))/(x) = 2.03 xx 10^(-4)` `x = (10^(-3))/(2.03 xx 10^(-4)) = 5 m` |
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