QJEEM24Ph3749: JEE Main 2024 Physics - Class 12 JEEM24Ph3749 Solution: Two wavelengths and are used in Young's double slit experiment. …

In Young's double slit experiment, the condition for constructive interference (bright fringes) is given by:

$d \sin \theta = n \lambda$

where:

• $d$ is the distance between the slits
• $\theta$ is the angle of the fringe relative to the central maximum
• $n$ is the order of the fringe (an integer)
• $\lambda$ is the wavelength of the light

We are given two different wavelengths:

$\lambda_1 = 450 \, \text{nm}$

$\lambda_2 = 650 \, \text{nm}$

For the fringes produced by these two wavelengths to overlap, the path difference must be an integer multiple of both wavelengths. This means:

$d \sin \theta = m \lambda_1 = n \lambda_2$

where $m$ and $n$ are the orders of the fringes for $\lambda_1$ and $\lambda_2$ , respectively.

To find the minimum order of fringe $n$ for $\lambda_2$ that coincides with a fringe for $\lambda_1$ , we need to find the least common multiple (LCM) of these wavelengths in terms of their smallest integers. This can be formulated as:

$m \lambda_1 = n \lambda_2$

Dividing both sides by $\lambda_1$ and $\lambda_2$ , we get:

$\frac{m}{\lambda_2} = \frac{n}{\lambda_1}$

Cross-multiplying, we get:

$m \lambda_1 = n \lambda_2$

Using the given wavelengths:

$m \times 450 = n \times 650$

Simplifying this equation, we get:

$\frac{m}{n} = \frac{650}{450}$

$\frac{m}{n} = \frac{13}{9}$

For the fringes to overlap, $m$ and $n$ must be integers. The smallest integers that satisfy this ratio are:

$m = 13$

$n = 9$

Therefore, the minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is:

$n = 9$