Question 25: Wave Optics

Question

In a single slit Fraunhofer diffraction pattern obtained at normal incidence, at the angular position of the second diffraction minimum the phase difference (in radian) between the waves from the opposite edges of the slit is

Accepted Answer

When two coherent waves of intensities $I_1$ and $I_2$ interfere, their extreme intensities are
Ã¢ÂÂ¢ Constructive (maximum):
$$I_{\max} = I_1 + I_2 + 2\sqrt{I_1\,I_2}$$
Ã¢ÂÂ¢ Destructive (minimum):
$$I_{\min} = I_1 + I_2 - 2\sqrt{I_1\,I_2}$$
Adding these gives
$$I_{\max} + I_{\min} = \bigl(I_1 + I_2 + 2\sqrt{I_1I_2}\bigr) + \bigl(I_1 + I_2 - 2\sqrt{I_1I_2}\bigr) = 2\,(I_1 + I_2)\,. $$
So the correct choice is
Option B: $\;2\,(I_1 + I_2)$.

Answer

$\pi$

Answer

zero

Answer

$2\pi$

Answer

$\frac{\pi}{2}$

Wave Optics

In a single slit Fraunhofer diffraction pattern obtained at normal incidence, at the angular position of the second diffraction minimum the phase difference (in radian) between the waves from the opposite edges of the slit is

Choose the correct answer: