First slide

Young's double slit Experiment

Question

Three waves of equal frequency having amplitudes $10 μm, 4 μm, 7 μm$ arrive at a given point with successive phase difference of $π / 2,$ the amplitude of the resulting wave in $μ m$ is given by

Moderate

A

4
B

5
C

6
D

7

Solution

The amplitudes of the waves are
$a_{1} = 10 μm, a_{2} = 4 μm and a_{3} = 7 μm$
and the phase difference between 1st and 2nd wave is
$\frac{π}{2}$ and that between 2nd and 3rd wave is $\frac{π}{2}$ .
Therefore, phase difference between 1st and 3rd is $π$ . Combining 1st with 3rd, their resultant amplitude is given by
$A_{1}^{2} = a_{1}^{2} + a_{3}^{2} + 2 a_{1} a_{3} cosϕ$
$or A_{1} = \sqrt{10^{2} + 7^{2} + 2 \times 10 \times 7 cosπ}$
$= \sqrt{100 + 49 - 140}$
$= \sqrt{9} = 3 μ m$ in the direction of first.
Now combining this with 2nd wave we have, the resultant amplitude
$A^{2} = A_{1}^{2} + a_{2}^{2} + 2 A_{1} a_{2} \cos \frac{π}{2}$
$or A = \sqrt{3^{2} + 4^{2} + 2 \times 3 \times 4 \cos 90^{\circ}} = \sqrt{9 + 16}$
$= 5 μ m$

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