Theoretical Foundation: Refractive Index and Dispersion

The interaction between light and a dielectric medium, such as glass, is governed by the frequency-dependent nature of the material's permittivity. The speed of light $v$ in a medium is defined by the relation:

$v = \frac{c}{n(\lambda)}$

where $c$ is the speed of light in a vacuum ($3 \times 10^8 \text{ m/s}$) and $n(\lambda)$ is the refractive index of the medium, which is a function of the wavelength $\lambda$ of the incident light [Per Cauchy's Dispersion Formula: $n(\lambda) = A + \frac{B}{\lambda^2} + \dots$].

Step 1: Addressing Independence of Speed

The speed of light in glass is not independent of the color of light. Because the refractive index $n$ of glass varies with the wavelength (frequency) of the incident radiation, different colors experience different phase velocities within the medium. This phenomenon is known as chromatic dispersion.

Step 2: Comparative Analysis of Red and Violet Light

To determine which color travels slower, we must evaluate the relationship between wavelength and refractive index:

• Wavelengths: The wavelength of violet light ($\lambda_v \approx 400 \text{ nm}$) is shorter than the wavelength of red light ($\lambda_r \approx 700 \text{ nm}$).
• Cauchy's Relation: As $\lambda$ decreases, $\frac{1}{\lambda^2}$ increases, leading to a higher value for $n$. Consequently, $n_{violet} > n_{red}$.
• Velocity Relation: Since $v \propto \frac{1}{n}$, a higher refractive index results in a lower velocity.

Step 3: Visual Representation of Dispersion

Data Comparison Table

Final Solution: The speed of light in glass is dependent on the color of light. Violet light has a shorter wavelength, resulting in a higher refractive index for the glass medium, which causes violet light to travel slower than red light.

Step 1: Establishing Physical Parameters

To determine the speed of light in a medium, we utilize the relationship between the refractive index of the medium, the speed of light in a vacuum, and the speed of light in the specified medium.

• Refractive index of glass ($\mu$ or $n$): $1.5$
• Speed of light in vacuum ($c$): $3.0 \times 10^8 \, \text{m s}^{-1}$
• Speed of light in glass ($v$): Unknown

Step 2: Defining the Governing Relation

By definition, the absolute refractive index of a medium is the ratio of the speed of light in a vacuum ($c$) to the speed of light in the medium ($v$). [Per the Definition of Refractive Index]:

$n = \frac{c}{v}$

Step 3: Algebraic Rearrangement and Calculation

We rearrange the equation to solve for the target variable $v$:

$v = \frac{c}{n}$

Substituting the given numerical values:

$v = \frac{3.0 \times 10^8 \, \text{m s}^{-1}}{1.5}$

Perform the division:

$v = 2.0 \times 10^8 \, \text{m s}^{-1}$

Visual Representation of Wave Refraction

Summary of Analytical Results

The calculation confirms that as light enters a medium with a higher refractive index ($n > 1$), its velocity decreases proportionately to the index of the material. Given $n=1.5$, the velocity is reduced by a factor of exactly 1.5.

Final Solution: The speed of light in glass is $2.0 \times 10^8 \, \text{m s}^{-1}$.

Theoretical Foundation: Wavefront Geometry

A wavefront is defined as the locus of all points having the same phase of oscillation. The shape of a wavefront is determined by the geometry of the source and the path traversed by the light rays, which propagate perpendicularly to the wavefront [Per Huygens-Fresnel Principle].

Step 1: Identifying the Source Geometry

The source is identified as a point source. A point source radiates light energy symmetrically in all directions in a homogeneous, isotropic medium.

Step 2: Defining the Path of Rays

Let the point source be located at the origin $O(0,0,0)$ in a three-dimensional coordinate system. Light rays travel radially outward from this source along all possible directions in space. Consequently, the distance $r$ from the source at any time $t$ for a given wavefront is $r = ct$, where $c$ is the speed of light.

Step 3: Geometrical Analysis

Since the rays travel an equal distance $r$ from the point source $O$ in all directions, the set of all points $(x, y, z)$ having the same phase satisfies the equation of a sphere:

$x^2 + y^2 + z^2 = r^2$

[By the definition of a sphere in 3D Euclidean geometry, where the locus of points is equidistant from a fixed center].

Analytical Conclusion

Because the rays are radial vectors diverging from the point source, the wavefront, being a surface perpendicular to these rays at all points, must necessarily be spherical.

Final Solution: The shape of the wavefront for light diverging from a point source is spherical.

Step 1: Establishing the Relationship between Path Difference and Phase Difference

The phase difference ($\phi$) is related to the path difference ($\Delta x$) by the following geometric relation:

$\phi = \frac{2\pi}{\lambda} \Delta x$ [Per the principle of wave superposition and phase-path relationship]

Step 2: Determining Intensity at Path Difference $\Delta x = \lambda$

Given $\Delta x_1 = \lambda$, we calculate the phase difference $\phi_1$:

$\phi_1 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$ radians.

The intensity ($I$) at any point on the screen is given by the formula:

$I = I_{max} \cos^2\left(\frac{\phi}{2}\right)$ [Based on the superposition of two coherent waves with equal amplitude]

Substituting $\phi_1 = 2\pi$:

$K = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max} (-1)^2 = I_{max}$

Thus, $I_{max} = K$.

Step 3: Calculating Intensity at Path Difference $\Delta x = \frac{\lambda}{3}$

For $\Delta x_2 = \frac{\lambda}{3}$, we calculate the phase difference $\phi_2$:

$\phi_2 = \frac{2\pi}{\lambda} \cdot \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$ radians.

Now, calculate the new intensity ($I'$) using the established $I_{max} = K$:

$I' = K \cos^2\left(\frac{\phi_2}{2}\right)$

$I' = K \cos^2\left(\frac{2\pi/3}{2}\right) = K \cos^2\left(\frac{\pi}{3}\right)$

Since $\cos(\pi/3) = 0.5$ or $\frac{1}{2}$:

$I' = K \left(\frac{1}{2}\right)^2 = \frac{K}{4} = 0.25K$

Visual Representation of Intensity Distribution

Final Solution: The intensity of light at a point where the path difference is $\frac{\lambda}{3}$ is $\frac{K}{4}$ units.

Given Variables & Initial Setup

The parameters provided for the Young’s Double-Slit Experiment (YDSE) are as follows:

• Slit separation ($d$): $0.28\text{ mm} = 0.28 \times 10^{-3}\text{ m}$
• Distance to the screen ($D$): $1.4\text{ m}$
• Distance of the fourth bright fringe ($y_4$): $1.2\text{ cm} = 1.2 \times 10^{-2}\text{ m}$
• Order of the fringe ($n$): $4$

Step 1: Establishing the Theoretical Foundation

In a YDSE, the position of the $n^{th}$ bright fringe from the central maximum is governed by the formula:
$y_n = \frac{n\lambda D}{d}$ [Per the condition for constructive interference]

Step 2: Algebraic Rearrangement

To isolate the wavelength ($\lambda$), we rearrange the formula as follows:
$\lambda = \frac{y_n \cdot d}{n \cdot D}$

Step 3: Numerical Substitution & Calculation

Substituting the provided values into the equation:

Step 4: Unit Conversion to Nanometers

Converting the wavelength to standard spectroscopic units (nanometers):
$\lambda = 6.0 \times 10^{-7} \times 10^9\text{ nm}$
$\lambda = 600\text{ nm}$

Visual Representation of Path Difference

Final Solution: The wavelength of the light used in the experiment is $600\text{ nm}$ (or $6 \times 10^{-7}\text{ m}$).

Given Parameters & Theoretical Foundation

In a Young’s Double-Slit Experiment (YDSE), the position of the $n$-th bright fringe for a wavelength $\lambda$ is given by the formula:

$y_n = \frac{n \lambda D}{d}$

Where:

• $\lambda_1 = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m}$
• $\lambda_2 = 520 \, \text{nm} = 520 \times 10^{-9} \, \text{m}$
• $D$ is the distance between the slits and the screen.
• $d$ is the separation between the two slits.

Step 1: Establishing the Condition for Coincidence

For the bright fringes of both wavelengths to coincide at a distance $y$ from the central maximum, the following condition must hold [Per the Principle of Superposition]:

$y = n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}$

Where $n_1$ and $n_2$ are the orders of the bright fringes for wavelengths $\lambda_1$ and $\lambda_2$ respectively.

Step 2: Solving for the Ratio of Orders

By equating the expressions and canceling the common geometric terms ($D/d$):

$n_1 \lambda_1 = n_2 \lambda_2$

$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650}$

Simplifying the fraction by dividing by the greatest common divisor ($130$):

$\frac{n_1}{n_2} = \frac{4}{5}$

Thus, the fourth bright fringe of $\lambda_1$ coincides with the fifth bright fringe of $\lambda_2$.

Step 3: Geometric Visualization of Interference

Step 4: Calculation of Least Distance

Using the value $n_1 = 4$ for $\lambda_1$:

$y = \frac{4 \times 650 \times 10^{-9} \times D}{d} = \frac{2600 \times 10^{-9} \times D}{d} \, \text{m}$

Using the value $n_2 = 5$ for $\lambda_2$ (verification):

$y = \frac{5 \times 520 \times 10^{-9} \times D}{d} = \frac{2600 \times 10^{-9} \times D}{d} \, \text{m}$

The distance is expressed as a multiple of the fringe width factor $\frac{D}{d}$:

$y = 2.6 \times 10^{-6} \frac{D}{d} \, \text{meters}$

Final Solution: The least distance from the central maximum where the bright fringes coincide is $2.6 \times 10^{-6} \frac{D}{d} \, \text{m}$.

Given Variables & Initial Setup

The parameters provided for the Young’s Double-Slit Experiment (YDSE) are as follows:

• Wavelength $\lambda_1 = 650\text{ nm} = 650 \times 10^{-9}\text{ m}$
• Wavelength $\lambda_2 = 520\text{ nm} = 520 \times 10^{-9}\text{ m}$
• Fringe order $n = 3$
• Variables $D$ (screen distance) and $d$ (slit separation) are constants for the experimental setup.

Step 1: Theoretical Foundation

In a YDSE setup, the position of the $n^{th}$ bright fringe (constructive interference) from the central maximum is governed by the formula derived from the path difference condition $\Delta x = n\lambda$:

$y_n = \frac{n \lambda D}{d}$

[Per the principle of superposition and the condition for constructive interference, where path difference must be an integral multiple of the wavelength].

Step 2: Calculating the Distance for $\lambda_1$

We are asked to find the distance of the $3^{rd}$ bright fringe ($n=3$) for light with wavelength $\lambda_1 = 650\text{ nm}$.

$y_3 = \frac{3 \times (650 \times 10^{-9}\text{ m}) \times D}{d}$

$y_3 = \frac{1950 \times 10^{-9} D}{d}\text{ m}$

$y_3 = 1.95 \times 10^{-6} \left(\frac{D}{d}\right)\text{ meters}$

Visual Representation of Fringe Geometry

Step 3: Analytical Conclusion

The distance is expressed as a function of the experimental ratio $\frac{D}{d}$. Given the standard structure of the problem, the distance $y_3$ is effectively $3\beta_1$, where $\beta_1$ is the fringe width for $\lambda_1$.

$\beta_1 = \frac{\lambda_1 D}{d} = 650 \times 10^{-9} \left(\frac{D}{d}\right)$

$y_3 = 3 \times 650 \times 10^{-9} \left(\frac{D}{d}\right) = 1.95 \times 10^{-6} \left(\frac{D}{d}\right)\text{ m}$

Final Solution: The distance of the third bright fringe from the central maximum is $1.95 \times 10^{-6} \frac{D}{d}\text{ meters}$.

Step 1: Analyzing the Source Characteristics

A star, while intrinsically a spherical source of light, is situated at an astronomical distance ($d \rightarrow \infty$) relative to an observer on Earth. In wave optics, the shape of a wavefront is determined by the locus of points having the same phase of oscillation [Per Huygens-Fresnel Principle].

Step 2: Geometric Transformation of Wavefronts

For any point source emitting light spherically, the wavefront at a radius $r$ is defined by a sphere with surface area $A = 4\pi r^2$. As the distance $r$ increases, the radius of curvature $R$ of the wavefront also increases. Mathematically, the curvature $\kappa$ is defined as:

$$\kappa = \frac{1}{R}$$

As $r \to \infty$, the radius of curvature $R \to \infty$, which implies that the curvature $\kappa \to 0$ [By the limit definition of geometric optics].

Step 3: Defining the Intercepted Wavefront

When a wavefront with an infinitely large radius of curvature is intercepted over a finite area (the portion of the Earth's surface or the aperture of an Earth-based telescope), the segment of the sphere is locally indistinguishable from a plane. This is analogous to the concept that a small arc on a circle of infinite radius behaves as a straight line.

Step 4: Conclusion based on Wave Theory

Since the light rays from a distant star are essentially parallel at the point of incidence on Earth, the phase fronts are perpendicular to these rays. A collection of parallel rays corresponds to a wavefront with an infinite radius of curvature [Per the Eikonal Approximation in optics].

Final Solution: The shape of the wavefront intercepted by the Earth from a distant star is a plane wavefront.