question_answer

A plane wave of wavelength 6250 A is incident normally on a slit of width \[2\times {{10}^{-2}}\]cm. The width of the principal maximum on a screen distant 50 cm will be

A) \[312.5\times {{10}^{-3}}\]cm  

B) \[312.5\times {{10}^{-6}}\]m

C) \[312.5\times {{10}^{-3}}\]m

D)        \[312.5\times {{10}^{-6}}\]cm

Correct Answer: A

Solution :

[a] Width of central maximum \[=\frac{2\,\lambda \,D}{a}=\frac{2\times 6250\times {{10}^{-10}}\times 0.5}{2\times {{10}^{-4}}}\] \[=3125\times {{10}^{-6}}m=312.5\times {{10}^{-3}}\,\,cm.\] (Screen position of various minima for Fraunh offer diffraction pattern of a single slit of width a).