A ray of light travelling in air is incident at grazing angle (Incident angle = 90 degrees) on a lon

Q: Why is the angle of incidence at the origin $90^\circ$?

Because the ray enters at a grazing angle, meaning it is parallel to the surface (x-axis). The normal is along the y-axis, so the angle with the normal is $90^\circ$.

Q: How do we know Snell's Law is valid for a continuously varying medium?

A continuously varying medium can be thought of as an infinite number of infinitesimally thin parallel layers. Snell's Law holds across each boundary, so $n \sin i$ remains constant throughout the entire medium.

Q: How did we get $\cot i$ from $\sin i$?

We used the trigonometric identity $\cot i = \sqrt{1 - \sin^2 i} / \sin i$. Alternatively, you can use $\csc^2 i - 1 = \cot^2 i$. Substituting the value of $\sin i$ gives the required expression.

Q: Why did we integrate from $0$ to $x$ and $0$ to $y$?

The ray starts at the origin $(0,0)$, so the initial conditions for the integration are $x=0$ when $y=0$.

Q: Why does the ray emerge grazingly?

By applying Snell's Law between the origin and the air outside the upper boundary: $n_{\text{origin}} \sin i_{\text{origin}} = n_{\text{air}} \sin e$. Since $n_{\text{origin}} = 1$, $i_{\text{origin}} = 90^\circ$, and $n_{\text{air}} = 1$, we get $\sin e = 1$, which means $e = 90^\circ$.

Question

A ray of light travelling in air is incident at grazing angle (Incident angle = $90^\circ$) on a long rectangular slab of a transparent medium of thickness $t = 1.0$ m. The point of incidence is the origin $A(0,0)$. The medium has a variable index of refraction $n(y)$ given by
$n(y) = [ky^{3/2} + 1]^{1/2}$ where $k = 1.0 	ext{ (m)}^{-3/2}$.
The refractive index of air is 1.0. (1995)
(a) Obtain a relation between the slope of the trajectory of the ray at a point $B(x, y)$ in the medium and the incident angle at that point.
(b) Obtain an equation for the trajectory $y(x)$ of the ray in the medium.
(c) Determine the coordinates $(x_1, y_1)$ of the point $P$, where the ray intersects the upper surface of the slab-air boundary.
(d) Indicate the path of the ray subsequently.

Sigma Percentile · Accepted Answer

### The Setup: A Variable Refractive Index

Imagine a light ray entering a glass slab. But this isn't an ordinary slab. Its refractive index changes continuously as we move along the y-axis, given by the function $n(y) = \sqrt{y^{3/2} + 1}$. The ray enters at a grazing incidence at the origin $A(0,0)$, meaning it is initially parallel to the x-axis. Our goal is to find the exact mathematical path this ray takes through the medium.

### Finding the Slope

Let's look at an arbitrary point $B(x,y)$ on the ray's path. The tangent to the path here makes an angle $	heta$ with the x-axis. The angle of incidence, $i$, is measured from the normal (which is parallel to the y-axis). From simple geometry, we can see that $	heta$ and $i$ are complementary, meaning $	heta = 90^\circ - i$.

The slope of the trajectory is given by the derivative $\frac{dy}{dx}$, which is equal to $	an 	heta$. Substituting our relation for $	heta$, we get:

$$\frac{dy}{dx} = 	an(90^\circ - i) = \cot i$$

This is our first crucial relation.

### Applying Snell's Law

Now, let's apply Snell's Law. In a continuously varying medium, the product of the refractive index and the sine of the angle of incidence remains constant along the entire path of the ray.

At the origin $A(0,0)$, the refractive index is $n(0) = 1$, and the angle of incidence is $90^\circ$ (grazing incidence). At point $B$, the refractive index is $n(y)$. Equating the two states:

$$n_A \sin i_A = n_B \sin i_B$$

$$1 	imes \sin 90^\circ = \sqrt{y^{3/2} + 1} \sin i$$

Since $\sin 90^\circ = 1$, we can easily isolate $\sin i$:

$$\sin i = \frac{1}{\sqrt{y^{3/2} + 1}}$$

Using basic trigonometric identities, we can find $\cot i$. We know that $\csc^2 i = y^{3/2} + 1$, and $\cot^2 i = \csc^2 i - 1$. This gives us $\cot^2 i = y^{3/2}$, which simplifies to:

$$\cot i = y^{3/4}$$

### The Differential Equation

Equating our two expressions for $\cot i$, we obtain a simple, separable differential equation:

$$\frac{dy}{dx} = y^{3/4}$$

Let's separate the variables and integrate from the origin $(0,0)$ to an arbitrary point $(x,y)$:

$$\int_0^y y^{-3/4} dy = \int_0^x dx$$

Integrating both sides yields:

$$\frac{y^{1/4}}{1/4} = x$$

$$4y^{1/4} = x$$

This is the exact equation of the ray's trajectory through the medium!

### The Point of Emergence

The ray emerges from the upper surface of the slab, where the thickness $y = 1.0$ m. Substituting $y = 1$ into our trajectory equation, we find $x = 4(1)^{1/4} = 4$ m. Therefore, the coordinates of the point of emergence $P$ are $(4.0 	ext{ m}, 1.0 	ext{ m})$.

Finally, what is the direction of the emergent ray? We can apply Snell's Law between the origin and the point of emergence into the air. Since the refractive index of air is 1 (the same as at the origin), the angle of emergence must be equal to the initial angle of incidence. Thus, the ray emerges at $90^\circ$ to the normal, meaning it grazes the upper surface of the slab.

Visualized Solution

Visualizing the Setup

Slope and Angle of Incidence

Relation for Slope

Applying Snell's Law

Substituting the Refractive Index

Finding $cot i$

Forming the Differential Equation

Solving for the Trajectory

Coordinates of Point P

Path of the Emergent Ray

The Sigma Insight: Refraction and Total Internal Reflection

The Setup

Finding the Slope

Applying Snell's Law

The Differential Equation

The Point of Emergence

Similar Questions

A ray of light $A O$ in vacuum is incident on a glass slab at angle $6 0^{\circ}$ and refracted at angle $3 0^{\circ}$ along $O B$ as shown in the figure. The optical path length of light ray from $A$ to $B$ is

Visualized Solution

Visualizing the Setup

Slope and Angle of Incidence

Relation for Slope

Applying Snell's Law

Substituting the Refractive Index

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } coti

Forming the Differential Equation

Solving for the Trajectory

Coordinates of Point P

Path of the Emergent Ray

The Sigma Insight: Refraction and Total Internal Reflection

The Setup

Finding the Slope

Applying Snell's Law

The Differential Equation

The Point of Emergence

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Finding $cot i$