A double slit interference experiment performed with a light of wavelength 600nm$600\mathrm{nm}$ forms an interference fringe pattern on a screen with 10th$10^{\mathrm{th}}$ bright fringe having its centre at a distance of 10mm$10\mathrm{mm}$ from the central maximum. Distance of the centre of the same 10th$10^{\mathrm{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660nm$660\mathrm{nm}$ would be ____________________

Related Formula

For two waves of identical direction and frequency superimposing with phase difference φ$\phi$:

Core Logic

The amplitudes of the two waves are E₁ = E₀$E_1 = E_0$ and E₂ = E₀$E_2 = E_0$.

The phase difference is:

Substitute these values into the resultant amplitude equation:

Step 1: Simplify the calculation

Since ((π)/(3)) = 0.5$\cos\left(\frac{\pi}{3}\right) = 0.5$:

This is closest to 1.7E₀$1.7\mathrm{E}_0$.

Pattern Recognition

Sees: Equal amplitudes (A$A$) with a phase angle of 60^°$60^\circ$ (π/3${\pi}/{3}$). Shortcut: The vector sum of two vectors of equal magnitude A$A$ separated by 60^°$60^\circ$ is always √(3)A ≈ 1.73A$\sqrt{3}A \approx 1.73A$. If separated by 120^°$120^\circ$, it is A$A$. If 90^°$90^\circ$, it is √(2)A$\sqrt{2}A$.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

For a stable, visible interference pattern to form, the light sources passing through the two slits must be coherent:

• They must have the same wavelength (or frequency).
• They must maintain a constant phase difference over time.

Core Logic

If one slit is covered by a red filter and the other by a green filter:

• Only red light (λred ≈ 700~nm$\lambda_{\text{red}} \approx 700\mathrm{~nm}$) passes through the first slit.
• Only green light (λgreen ≈ 500~nm$\lambda_{\text{green}} \approx 500\mathrm{~nm}$) passes through the second slit.

Since the two passing waves have completely different wavelengths and frequencies, they are incoherent.

Step 1: Resulting Pattern Analysis

Coherent sources are a prerequisite for producing stable bright and dark interference fringes. Incoherent waves of different frequencies merely superimpose to create a general background illumination without any distinct, observable spatial fringe lines.

Thus, there shall be no interference fringes.

Pattern Recognition

Sees: YDSE + opposite colored filters (red and green) on slits. Trap: Don't get confused thinking separate patterns will overlay. Since the slits emit different colors, the sources are incoherent, so the interference term φ = 0$\langle \cos\phi \rangle = 0$. Shortcut: Different colors = different wavelengths = incoherent sources = NO fringes. ✓

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

For superposition of two coherent beams of intensities I₁$I_1$ and I₂$I_2$:

Core Logic

Given values:

• I₁ = 4I$I_1 = 4I$
• I₂ = 9I$I_2 = 9I$

Let's calculate the square roots of the intensities:

• √(I₁) = √(4I) = 2√(I)$\sqrt{I_1} = \sqrt{4I} = 2\sqrt{I}$
• √(I₂) = √(9I) = 3√(I)$\sqrt{I_2} = \sqrt{9I} = 3\sqrt{I}$

Step 1: Calculating Max and Min Intensities

Substitute these values into the intensity formulas:

Step 2: Finding the Difference

The difference between the maximum and minimum intensities is:

Since this difference is given as xI$xI$:

x = 24$x = 24$

Pattern Recognition

Algebraic Shortcut:

Substitute I₁ = 4I$I_1 = 4I$ and I₂ = 9I$I_2 = 9I$: 4√(4I · 9I) = 4√(36 I²) = 4 × 6I = 24I$$4\sqrt{4I \cdot 9I} = 4\sqrt{36 I^2} = 4 \times 6I = 24I$$. This beautiful identity (4ab$4ab$ formula) lets you solve the problem instantly without separately calculating maximum and minimum values!

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Malus's Law:

Core Logic

Since P₁$P_1$ and P₂$P_2$ are crossed, the angle between their transmission axes is 90^°$90^\circ$. Let the angle between P₁$P_1$ and P₃$P_3$ be θ$\theta$. Then the angle between P₃$P_3$ and P₂$P_2$ is (90^° - θ)$\left(90^\circ - \theta\right)$. Intensity after passing through P₃$P_3$: I₁ = I₀ ²θ$I_1 = I_0 \cos^2\theta$. Intensity after passing through P₂$P_2$: Iₙₑₜ = I₁ ²(90^° - θ) = I₀ ²θ ²θ$I_{\text{net}} = I_1 \cos^2(90^\circ - \theta) = I_0 \cos^2\theta \sin^2\theta$.

Step 1: Maximize Net Intensity

Rewrite the expression:

For maximum transmitted intensity, (2θ) = 1 2θ = 90^° θ = 45^° = (π)/(4)$\sin(2\theta) = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ = \frac{\pi}{4}$. The angle between P₂$P_2$ and P₃$P_3$ is 90^° - 45^° = 45^° = (π)/(4)$90^\circ - 45^\circ = 45^\circ = \frac{\pi}{4}$.

Pattern Recognition

Inserting a polarization filter at exactly 45^°$45^\circ$ (π/4$ \pi/4$) between crossed polarizers symmetrically splits up components, maximizing overall transmission throughput.

Chapter Mix

Class 12 Physics: Wave Optics