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:

$$n = 9$$