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 a thin film, the condition for minimum transmission (which corresponds to maximum reflection in a non-absorbing medium) satisfies consecutive destructive wave path interference bounds [cite: 846, 847]:

The volumetric rate of evaporation from the circular boundary surface area is given by:

Core Logic

Given parameters:

• Refractive index, μ = 1.4$\mu = 1.4$
• Ring radius, R = 1.8 cm = 1.8 × 10⁻² m$R = 1.8 \text{ cm} = 1.8 \times 10^{-2} \text{ m}$
• Wavelength, λ = 560 nm = 560 × 10⁻⁹ m$\lambda = 560 \text{ nm} = 560 \times 10^{-9} \text{ m}$
• Time interval between consecutive minima, t = 12 s$t = 12 \text{ s}$

Calculate the thickness change Δ t$\Delta t$ corresponding to consecutive transmission minima :

Now, compute the volume change over this 12-second window to find the volumetric evaporation rate :

Comparing this to the given expression π × 10⁻¹³ m³/s$\pi \times 10^{-13} \text{ m}^3/\text{s}$ identifies the coefficient[cite: 207, 843]:

Pattern Recognition

A minimum in transmission means maximum reflection. For thin-film interference, the optical path difference changes by exactly (λ)/(2)$\frac{\lambda}{2}$ between consecutive fringes, which corresponds to a physical thickness change of (λ)/(2μ)$\frac{\lambda}{2\mu}$.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Linear width of the central maximum in single slit diffraction:

where D$D$ corresponds directly to the focal length f$f$ of the focus lens.

Core Logic

Given configuration values: λ = 6000 AA = 6 × 10⁻⁷~m$\lambda = 6000\text{ \AA} = 6 \times 10^{-7}\mathrm{~m}$ a = 0.01~mm = 1 × 10⁻⁵~m$a = 0.01\mathrm{~mm} = 1 \times 10^{-5}\mathrm{~m}$ D = f = 20~cm = 0.2~m$D = f = 20\mathrm{~cm} = 0.2\mathrm{~m}$

Substitute parameter matrices:

Step 1: Simplify Scientific Notation

Pattern Recognition

When a convex lens is placed right after the slit, the distance D$D$ to the screen is exactly equal to the focal length f$f$ of the lens.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Phase difference φ$\phi$ in terms of path difference Δ x$\Delta x$:

Intensity of interference fringe at a point with phase difference φ$\phi$:

Core Logic

Given path difference:

Calculate phase difference φ$\phi$:

Step 1: Calculate the Intensity Ratio

Using the intensity formula:

Using trigonometric identity (2π - θ) = θ$\cos(2\pi - \theta) = \cos\theta$:

Since ((π)/(4)) = 1√(2)$\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:

Thus, the intensity ratio is 1/2$1/2$.

Pattern Recognition

Any path difference that is an odd multiple of (λ)/(4)$\frac{\lambda}{4}$ (like (7λ)/(4)$\frac{7\lambda}{4}$) corresponds to a phase difference of an odd multiple of (π)/(2)$\frac{\pi}{2}$ when halved. Consequently, ²$\cos^2$ of this value will always yield exactly (1)/(2)$\frac{1}{2}$.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

For single slit diffraction, the position of the n$n$-th minima on the screen is given by:

where:

• λ$\lambda$ is the wavelength of light.
• D$D$ is the distance to the screen.
• b$b$ is the width of the slit.

Core Logic

Given parameters:

• Wavelength, λ = 6000 AA = 6000 × 10⁻¹⁰ m$\lambda = 6000\text{ \AA} = 6000 \times 10^{-10}\text{ m}$
• Screen distance, D = 50 cm = 0.5 m$D = 50\text{ cm} = 0.5\text{ m}$
• Minima distance, y₃ - y₁ = 3 mm = 3 × 10⁻³ m$y_3 - y_1 = 3\text{ mm} = 3 \times 10^{-3}\text{ m}$

Step 1: Calculate Slit Width

Using the position formula, calculate the separation between 3rd and 1st minima:

Rearranging to solve for slit width b$b$:

Substitute the values:

Comparing this with x × 10⁻⁴ m$x \times 10^{-4}\text{ m}$, we get:

x = 2$x = 2$

Pattern Recognition

Distance between n$n$-th and m$m$-th minima is Δ y = (n-m)(λ D)/(b)$\Delta y = (n-m)\frac{\lambda D}{b}$. Since 3-1 = 2$3-1 = 2$, we get Δ y = 2(λ D)/(b)$\Delta y = 2\frac{\lambda D}{b}$. Substitute and scale variables directly.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

For a first order dark fringe (minima) to form, the net optical path difference between the interfering paths must be an odd multiple of half-wavelength:

Core Logic

From the symmetric setup diagram geometry:

• Total path length 1 = Source arrow S₁ arrow P = √(D² + d²) + √(D² + d²) = 2√(D² + d²)$\text{Source} \rightarrow S_1 \rightarrow P = \sqrt{D^2 + d^2} + \sqrt{D^2 + d^2} = 2\sqrt{D^2 + d^2}$
• Total path length 2 = Source arrow S₂ arrow P = D + D = 2D$\text{Source} \rightarrow S_2 \rightarrow P = D + D = 2D$

Hence, the exact geometric path difference is:

Step 1: Set up Path Difference Balance

Equating path difference to the minima criterion:

Step 2: Expand and Isolate Slit Space d

Squaring both sides:

Step 3: Approximate and Substitute Values

Since λ$\lambda$ (400 ~nm$400 \mathrm{~nm}$) is tiny compared to D$D$ (0.2 ~m$0.2 \mathrm{~m}$), the term (λ²)/(16)$\frac{\lambda^2}{16}$ is completely negligible:

Substituting values:

Therefore, the minimum distance between the slits is 0.20 ~mm$0.20 \mathrm{~mm}$.

Pattern Recognition

This non-standard double slit setup calculates the primary total path directly from geometric lines rather than relying on standard angular small angle assumptions (y d / D$y d / D$). Recognizing when secondary higher-order differentials (λ²$\lambda^2$) can be safely dropped keeps calculations clean.

Chapter Mix

Class 12 Physics: Wave Optics