In a Young’s double slit experiment, the intensities at two points, for the path difference 4 and 3 (where is the wavelength of light used), are I1 and I2 respectively. If I0 denotes the intensity…

Given: Path differences are 4 and 3. The intensity due to each individual slit is I0. Find: The value of I1 + I2I0 For equal slit intensities in YDSE, the resultant intensity is I = 4I0 ^2 ( 2) For path difference 4, = 2 4 = 2 So, I1 = 4I0 ^2 ( 4) = 2I0 For path difference 3, = 2 3 = 2 3 So, I2 = 4I0 ^2 ( 3) = I0 Therefore, I1 + I2I0 = 2I0 + I0I0 = 3 The required numerical value is 3.

In a Young’s double slit experiment, the intensities at two point… — Solution | JEE Physics

NVAMediumJEE 2023Interference (Young's Experiment)

In a Young’s double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ (where $\lambda$ is the wavelength of light used), are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each of the individual slits, then

\frac{I_1 + I_2}{I_0} = \ldots

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: Path differences are $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ . The intensity due to each individual slit is $I_0$ .

Find: The value of

\frac{I_1 + I_2}{I_0}

For equal slit intensities in YDSE, the resultant intensity is

I = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right)

For path difference $\frac{\lambda}{4}$ ,

\Delta \phi = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{4} = \frac{\pi}{2}

So,

I_1 = 4I_0 \cos^2\left(\frac{\pi}{4}\right) = 2I_0

For path difference $\frac{\lambda}{3}$ ,

\Delta \phi = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{3} = \frac{2\pi}{3}

So,

I_2 = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = I_0

Therefore,

\frac{I_1 + I_2}{I_0} = \frac{2I_0 + I_0}{I_0} = 3

The required numerical value is $3$ .

Common mistakes

Using path difference directly inside cosine without converting it to phase difference is incorrect. First use $\Delta \phi = \frac{2\pi}{\lambda}\Delta x$ , then substitute into the intensity formula.
Taking the interference formula as $I = 2I_0\cos^2\left(\frac{\Delta\phi}{2}\right)$ is wrong here. Since each slit individually produces intensity $I_0$ , the combined maximum form becomes $I = 4I_0\cos^2\left(\frac{\Delta\phi}{2}\right)$ .
Forgetting that $\cos^2\left(\frac{\pi}{4}\right)=\frac{1}{2}$ and $\cos^2\left(\frac{\pi}{3}\right)=\frac{1}{4}$ leads to wrong intensities. Evaluate the trigonometric factors carefully before multiplying by $4I_0$ .

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