Interference (Ydse): Comprehensive NEET Physics Notes

1. Interference in Young’s Double Slit Experiment (YDSE)

1.1 Introduction to Interference

Interference occurs when two or more light waves overlap, resulting in regions of increased or decreased intensity. This phenomenon demonstrates the wave nature of light and is evident in the Young's Double Slit Experiment (YDSE).

• Constructive Interference: Occurs when waves overlap in phase (crest to crest or trough to trough), leading to increased brightness.
• Destructive Interference: Occurs when waves overlap out of phase (crest to trough), resulting in darkness.

1.2 Young’s Double Slit Experiment (YDSE)

The YDSE is a classical experiment demonstrating the wave nature of light through interference patterns.

Experimental Setup

• Two slits, $S_1$ and $S_2$, are illuminated by a single monochromatic light source, ensuring coherent light waves emanate from both slits.
• The light waves from these slits interfere on a screen placed at a distance, creating an alternating pattern of bright and dark fringes.

Path Difference and Fringe Formation

• Path Difference ($\Delta x$): The difference in distance traveled by light waves from the two slits to any point on the screen.
• • Constructive interference occurs when $\Delta x=n\lambda$, where $n$ is an integer and $\lambda$ is the wavelength of light.
  • Destructive interference occurs when $\Delta x=(n+\frac{1}{2})\lambda$.

Fringe Width and Fringe Spacing

• Fringe Width (β): The distance between two successive bright or dark fringes.
  $\beta =\frac{\lambda D}{d}$
  Where:
• • $\lambda$ = Wavelength of light
  • $D$ = Distance between the screen and slits
  • $d$ = Distance between the two slits

Did You Know?
Thomas Young's experiment in 1801 was the first to confirm the wave nature of light, providing the foundation for wave optics.

1.3 Mathematical Analysis of YDSE

Bright Fringes

The position of bright fringes (maxima) is given by: $x_n=\frac{n\lambda D}{d}$

Dark Fringes

The position of dark fringes (minima) is given by: $x_n=\frac{(n+\frac{1}{2})\lambda D}{d}$

Real-life Application:
Interference patterns are utilized in technologies such as interferometers, which measure tiny distances or changes in the refractive index of materials with high precision.

1.4 Coherence in YDSE

For visible interference, the light sources must be coherent, meaning they emit light waves with a constant phase difference and the same wavelength.

Conditions for Coherent Interference

1. The sources must emit monochromatic light (same wavelength).
2. They must maintain a constant phase difference.
3. They should have the same or nearly identical amplitude.

Common Misconception:
It's often believed that any two light sources can produce interference. In reality, without coherence, an interference pattern cannot be observed.

1.5 Intensity Distribution in YDSE

The resultant intensity $I$ at any point is related to the intensity of individual waves ($I_0$) as:

$I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

Where $\varphi$ represents the phase difference between the two waves.

NEET Problem-Solving Strategy:
Always identify whether the question involves constructive or destructive interference. Apply the corresponding path difference formulas accurately.

Quick Recap

• Interference results from the superposition of two or more light waves.
• Bright fringes occur when $x=\frac{n\lambda D}{d}$ and dark fringes when $x=\frac{(n+\frac{1}{2})\lambda D}{d}$.
• Fringe width is determined by $\beta =\frac{\lambda D}{d}$.
• Coherent light sources are essential for clear interference patterns.

Concept Connection

Link to Chemistry: The interference of light waves is similar to the constructive and destructive overlap of electron orbitals in molecules, explaining bonding and antibonding interactions in chemistry.

Practice Questions

Question 1

In a YDSE setup, two slits are separated by 0.2 mm, and the screen is 1 m away. The light used has a wavelength of 600 nm. Calculate the fringe width.

Solution: $\beta =\frac{\lambda D}{d}=\frac{600\times 10^{-9}\times 1}{0.2\times 10^{-3}}=3\times 10^{-3}m=3mm$

Question 2

Find the path difference at the second bright fringe if slit separation is 0.5 mm, screen distance is 1 m, and wavelength is 500 nm.

Solution: For the second bright fringe, $n=2$: $\Delta x=n\lambda =2\times 500\times 10^{-9}=1\times 10^{-6}m=1\mu m$

Question 3

If the intensity at the central bright fringe is $I_0$, what will be the intensity at a point where the path difference is $\frac{\lambda }{2}$?

Solution: Path difference $=\frac{\lambda }{2}$ corresponds to destructive interference, hence the intensity $I=0$.

Question 4

Determine the distance from the central maximum to the third dark fringe in a YDSE experiment with slit separation of 0.1 mm, screen distance of 1.5 m, and wavelength 600 nm.

Solution: For the third dark fringe, $n=2$: $x=\frac{(n+\frac{1}{2})\lambda D}{d}=\frac{(2.5)\times 600\times 10^{-9}\times 1.5}{0.1\times 10^{-3}}=22.5\times 10^{-3}m=22.5mm$

Question 5

Why is there no interference pattern when two independent sodium lamps illuminate two pinholes in YDSE?

Solution: Two independent sodium lamps are incoherent sources, meaning they do not maintain a constant phase difference, hence no interference pattern forms.

Visual Aids

1. Diagram of YDSE Setup: Include a detailed diagram showing the light source, slits, screen, and resulting interference pattern.
2. Interference Pattern: Provide a graphical representation of the bright and dark fringes to visualize constructive and destructive interference.
3. Wavefront Representation: Display how wavefronts from slits overlap and interfere.

Quick Reference Guide & Glossary

NEET Exam Strategy:
Focus on understanding the derivations and relations for constructive and destructive interference, as NEET frequently tests application-based understanding of these concepts.