A) 12.2 m
B) 24.2 m
C) 2.44 m
D) 1.22 m
Correct Answer: C
Solution :
According to Rayleigh's criterion, \[\theta =\frac{1.22\lambda }{{{d}_{e}}}\] where \[\lambda =\] wavelength of light, \[{{d}_{e}}=\] diameter of the pupil of the eye \[\therefore \] \[\theta =\frac{1.22\times 500\times {{10}^{-9}}}{2.5\times {{10}^{-3}}}=2.44\times {{10}^{-4}}\]radian But \[\theta =\frac{a}{D}\] \[\therefore \] Distance of separation, \[a=D\times \theta =10\times {{10}^{3}}\times 2.44\times {{10}^{-4}}\] \[=2.44\,m\]You need to login to perform this action.
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