In Young’s double slit experiment, light from two identical sourc… — Solution | JEE Physics

MCQEasyJEE 2024Interference (Young's Experiment)

In Young’s double slit experiment, light from two identical sources is superimposing on a screen. The path difference between the two lights reaching a point on the screen is $\frac{7\lambda}{4}$ . The ratio of intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

A
$\frac{1}{2}$
B
$\frac{3}{4}$
C
$\frac{1}{3}$
D
$\frac{1}{4}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Path difference $\Delta x = \frac{7\lambda}{4}$ .

Find: The ratio $\frac{I}{I_{\text{max}}}$ at this point.

The phase difference is related to path difference by

\phi = \frac{2\pi}{\lambda} \Delta x

Substituting $\Delta x = \frac{7\lambda}{4}$ ,

\phi = \frac{2\pi}{\lambda} \times \frac{7\lambda}{4} = \frac{7\pi}{2}

For two identical sources in Young’s double slit experiment,

I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right)

Hence,

\frac{I}{I_{\text{max}}} = \cos^2\left(\frac{\phi}{2}\right) = \cos^2\left(\frac{7\pi}{4}\right)

Now,

\cos\left(\frac{7\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

So,

\frac{I}{I_{\text{max}}} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}

Therefore, the ratio of intensity to maximum intensity is $\frac{1}{2}$ . The correct option is A.

Using Interference Intensity Relation

Given: In Young’s double slit experiment, the path difference is $\frac{7\lambda}{4}$ .

Find: The fringe intensity relative to the maximum intensity.

The intensity depends on the phase difference produced by the path difference. First convert path difference into phase difference using

\phi = \frac{2\pi}{\lambda} \times \Delta x

Substitute the given value:

\phi = \frac{2\pi}{\lambda} \times \frac{7\lambda}{4} = \frac{7\pi}{2}

Therefore,

\frac{\phi}{2} = \frac{7\pi}{4}

Now apply the relation

\frac{I}{I_{\text{max}}} = \cos^2\left(\frac{\phi}{2}\right)

So,

\frac{I}{I_{\text{max}}} = \cos^2\left(\frac{7\pi}{4}\right)

Using periodicity of cosine,

\cos\left(\frac{7\pi}{4}\right) = \cos\left(2\pi - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)

Hence,

\frac{I}{I_{\text{max}}} = \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}

Therefore, the required ratio is $\frac{1}{2}$ , so the correct option is A.

Common mistakes

Using the path difference directly in the cosine without converting it to phase difference is incorrect. First use $\phi = \frac{2\pi}{\lambda}\Delta x$ , then substitute into the intensity relation.
Confusing the formulas for amplitude and intensity leads to error. Intensity is proportional to the square of the resultant amplitude, so the correct ratio is $\cos^2\left(\frac{\phi}{2}\right)$ , not just $\cos\left(\frac{\phi}{2}\right)$ .
Taking $\cos\left(\frac{7\pi}{4}\right)$ with the wrong sign is a common mistake. Since $\frac{7\pi}{4} = 2\pi - \frac{\pi}{4}$ lies in the fourth quadrant, cosine is positive, giving $\frac{1}{\sqrt{2}}$ .

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