JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 19)
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are $$\pi / 2$$ and $$\pi / 3$$ at points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. The difference between the resultant intensities at the two points is $$x I$$. The value of $$x$$ will be ________.
Answer
2
Explanation
$${I_{{R_1}}} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $$
$${I_A} = I + 4I + 2\sqrt {I.4I} \cos 90^\circ $$
$$ = 5I$$
$${I_B} = I + 4I + 2\sqrt {I.4I} \cos 60^\circ $$
$$ = 7I$$
$${I_B} - {I_A} = 2I$$
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