Wave OpticsHard
Question
Light of wavelength 589.3 nm is incident normally on a slit of width 0.01mm. The angular width of the central diffraction maximum at a distance of 1m from the slit, its:
Options
A.0.68o
B.0.34o
C.2.05o
D.none of these
Solution
In case of diffraction,
Angular width of central fringe =
= 2 × 589.3 × 10-4
= 1178.6 × 10-5 rad.
= 1178.6 × 10-5 ×
= 67563 × 10-5
= 0.68o
Angular width of central fringe =
= 2 × 589.3 × 10-4= 1178.6 × 10-5 rad.
= 1178.6 × 10-5 ×
= 67563 × 10-5= 0.68o
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