Home › Physics › Q15
PhysicsWave Optics
In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ, is K units. The intensity of light at a point where the path difference is λ/3 will be :
Options
1
K/4
2
K
3
2K
4
K/4
Correct Answer
Option 1 : K/4
Step-by-Step Solution
1

Path difference λ → Phase difference:

$$\phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \lambda = 2\pi$$

Intensity: \(I = 4I_0\cos^2\!\left(\dfrac{2\pi}{2}\right) = 4I_0\cos^2\pi = 4I_0(1) = 4I_0 = K\)

So \(I_0 = K/4\)

2

Path difference λ/3 → Phase difference:

$$\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3}$$

3

Apply intensity formula:

$$I = 4I_0\cos^2\!\left(\frac{\pi}{3}\right) = 4I_0 \times \left(\frac{1}{2}\right)^2 = 4I_0 \times \frac{1}{4} = I_0 = \frac{K}{4}$$

Answer: K/4 ✓

Theory: Wave Optics & YDSE
1. Intensity Formula in YDSE

When two coherent light waves of equal intensity I₀ superpose, the resultant intensity depends on their phase difference φ:

\(I = 4I_0\cos^2\!\left(\dfrac{\phi}{2}\right)\)

where \(\phi = \dfrac{2\pi}{\lambda}\Delta x\) and Δx = path difference

At φ = 0 or 2π (path diff = 0 or λ): I = 4I₀ = I_max (bright fringe). At φ = π (path diff = λ/2): I = 0 = I_min (dark fringe). The average intensity = 2I₀, confirming energy conservation.

2. Fringe Width and Fringe Pattern

Fringe width β = λD/d, where D is screen distance and d is slit separation. All fringes have equal width. Bright fringes at path diff = nλ. Dark fringes at path diff = (2n−1)λ/2. The central bright fringe is at zero path difference.

📌 Δx = λ → φ = 2π → I = 4I₀ = K (maximum, bright fringe)

📌 Δx = λ/2 → φ = π → I = 0 (dark fringe)

📌 Δx = λ/3 → φ = 2π/3 → I = K/4

📌 Δx = λ/4 → φ = π/2 → I = 2I₀ = K/2

3. Conditions for Sustained Interference

For a stable interference pattern: (1) Sources must be coherent — same frequency and constant phase difference. (2) Sources must be monochromatic. (3) Amplitudes should be comparable (equal amplitudes give best contrast with I_min = 0). (4) Sources must be very close and screen far away. In YDSE, the two slits act as coherent sources derived from the same primary source.

4. Effect of Different Colours and Media

If wavelength changes (different colour), fringe width changes proportionally (β ∝ λ). In a medium of refractive index μ, effective wavelength = λ/μ, so fringe width decreases by factor μ. If entire apparatus is immersed in water (μ = 4/3), fringe width becomes 3/4 of air value. If white light is used, central fringe is white but higher order fringes show colour dispersion.

5. Common NEET Traps in YDSE

⚠️ cos²(φ/2) NOT cos²(φ) — always halve the phase before applying cosine

⚠️ Path diff λ gives constructive interference just like path diff = 0

⚠️ cos(60°) = 1/2, so cos²(60°) = 1/4 — memorise key cosine values

⚠️ I_max = 4I₀ only when both slits have equal intensity I₀

Frequently Asked Questions
1. How do we convert path difference to phase difference? ⌄
φ = (2π/λ) × Δx. For Δx = λ/3: φ = 2π/3. This is the fundamental relation connecting the geometric path difference to the wave phase difference.
2. Why does path difference λ give maximum intensity? ⌄
Path difference λ → phase difference 2π → the two waves are in phase (same as zero path difference). Constructive interference gives I_max = 4I₀.
3. What is the value of cos(π/3)? ⌄
cos(π/3) = cos(60°) = 1/2. Therefore cos²(π/3) = 1/4. This is a trigonometry value that must be memorised for NEET optics problems.
4. What is I₀ in terms of K? ⌄
Since 4I₀ = K (intensity at path diff = λ which is maximum), I₀ = K/4. Here I₀ is the intensity from each individual slit when the other is blocked.
5. What is fringe width formula? ⌄
β = λD/d, where λ = wavelength, D = distance from slits to screen, d = separation between slits. Fringe width is directly proportional to wavelength and screen distance, inversely proportional to slit separation.
6. What happens to fringe pattern if one slit is closed? ⌄
If one slit is closed, there is no interference. The screen shows a single-slit diffraction pattern — a broad central maximum with secondary maxima of decreasing intensity. Fringe pattern disappears completely.
7. What is coherence and why is it needed? ⌄
Coherence means the two sources maintain a constant phase relationship over time. Without coherence, the phase difference fluctuates randomly and the interference pattern averages out, leaving uniform illumination. Laser light is highly coherent; ordinary bulb light is not.
8. If slit separation d is doubled, what happens to fringe width? ⌄
β = λD/d. If d doubles, β halves — fringes become narrower. More fringes are visible in the same screen area. Conversely, bringing slits closer gives wider, more easily observable fringes.
9. What is the intensity at the central bright fringe? ⌄
At central fringe, path difference = 0, phase difference = 0. I = 4I₀cos²(0) = 4I₀×1 = 4I₀ = K. Maximum intensity — same as at path difference λ, 2λ, 3λ, etc.
10. What is the ratio I_max/I_min when both slits have equal intensity? ⌄
I_max = 4I₀, I_min = 0. Ratio = ∞. When slits have unequal intensities I₁ and I₂: I_max/I_min = (√I₁+√I₂)²/(√I₁−√I₂)².
Previous Questions
Q.
Sum of KE and PE of simple pendulum bob is 0.02 J – speed at equilibrium position
Oscillations · Answer: 1.41 m/s
Q.
Work done to raise mass m from Earth surface to height equal to radius R
Gravitation · Answer: mgR/2
Q.
Box of mass 15 kg on trolley, static friction 0.12 – maximum acceleration of trolley
Laws of Motion · Answer: 1.2 m/s²
Q.
Hydrogen atom first excited state energy −3.4 eV – radial distance from nucleus
Atoms & Nuclei · Answer: 2.1 × 10⁻¹⁰ m
Q.
Match List – E=hv, Diffraction, λ=h/p, Compton effect with dual nature properties
Dual Nature · Answer: A-IV, B-III, C-I, D-II