In Young's double slit experiment, width of the second slit is double the width of first slit, consequently the amplitude of the light from two slits. ' I m ' is the maximum intensity

Question

In Young's double slit experiment, width of the second slit is double the width of first slit, consequently the amplitude of the light from two slits. ' I m ​ ' is the maximum intensity. The resultant intensity ' I ' when they interfere with the phase difference of ϕ is given by

TestHub · Accepted Answer

Based on the given conditions, the amplitude of light from the second slit is twice that of the first slit. Let $A_{1} = A$ , then $A_{2} = 2 A$ .

The corresponding intensities are proportional to the square of the amplitudes, so $I_{1} = I_{0}$ and $I_{2} = 4 I_{0}$ .

The maximum intensity $I_{m}$ occurs when the phase difference is zero:

$I_{m} = (I_{1} + I_{2})^{2}$

$I_{m} = (I_{0} + 4 I_{0})^{2} = (3 I_{0})^{2} = 9 I_{0}$

$\Rightarrow I_{0} = \frac{I _{m}}{9}$

The resultant intensity $I$ at a phase difference $ϕ$ is given by:

$I = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ$

$I = I_{0} + 4 I_{0} + 2 4 I_{0}^{2} cos ϕ$

$I = 5 I_{0} + 4 I_{0} cos ϕ$

Using the trigonometric identity $cos ϕ = 2 cos^{2} \frac{ϕ}{2} - 1$ :

$I = 5 I_{0} + 4 I_{0} (2 cos^{2} \frac{ϕ}{2} - 1)$

$I = 5 I_{0} + 8 I_{0} cos^{2} \frac{ϕ}{2} - 4 I_{0}$

$I = I_{0} (1 + 8 cos^{2} \frac{ϕ}{2})$

Substituting $I_{0} = \frac{I _{m}}{9}$ :

$I = \frac{I _{m}}{9} (1 + 8 cos^{2} \frac{ϕ}{2})$

Answer: $\frac{I _{m}}{9} (1 + 8 cos^{2} \frac{ϕ}{2})$

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