JEE Main 2024 — Wave Optics Question with Solution

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:

$$n=9$$