In single slit fraunhoffer diffraction pattern, resultant amplitude at the centre of principal maxima is $A_{0}$ and it is $A$ at an angular position of 30°. Then $\frac{A}{A_{0}}$ will be __
[slit width $a = λ$ ]

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$\frac{π}{4}$

$\frac{2}{π}$

$\frac{π}{2}$

$\frac{4}{π}$

answer is B.

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Detailed Solution

$A = A_{0} (\frac{\sin β}{β}) o r \frac{A}{A_{0}} = \frac{\sin β}{β} N o w β = \frac{π a \sin θ}{λ} = \frac{π}{2} ∴ \frac{A}{A_{0}} = \frac{\sin (\frac{π}{2})}{(\frac{π}{2})} = \frac{2}{π}$

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