To determine how the width of a ray changes as it passes through a layer of variable refractive index and then into glass, we need to analyze the situation step by step. The problem involves a layer of refractive index that varies with the parameter \( y \), and we need to consider how light behaves at the boundaries of these different media.
Understanding the Refractive Index
The refractive index \( \mu \) of the variable layer is given by the equation:
\( \mu = \frac{2 - y}{\sqrt{2}} \)
This indicates that as \( y \) varies from 0 to 1, the refractive index changes from \( \mu(0) = \sqrt{2} \) (when \( y = 0 \)) to \( \mu(1) = \frac{1}{\sqrt{2}} \) (when \( y = 1 \)).
Incident Ray Characteristics
The ray of light is incident at an angle of 45 degrees in air, where the refractive index is approximately 1. The width of the ray is denoted as \( w \). When the ray enters the variable refractive index layer, it will undergo refraction according to Snell's Law:
\( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
Here, \( n_1 = 1 \) (air), \( \theta_1 = 45^\circ \), and \( n_2 = \mu \) (the variable refractive index). We will analyze the ray's behavior at both boundaries: air to the variable layer and then from the variable layer to glass.
Refraction at the Air to Variable Layer Boundary
Using Snell's Law at the boundary between air and the variable refractive index layer:
\( 1 \cdot \sin(45^\circ) = \mu \cdot \sin(\theta_2) \)
Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), we can express this as:
\( \frac{\sqrt{2}}{2} = \mu \cdot \sin(\theta_2) \)
Solving for \( \sin(\theta_2) \) gives:
\( \sin(\theta_2) = \frac{\sqrt{2}}{2\mu} \)
Calculating the Width Change
The width of the ray will change as it enters the variable layer due to the change in angle. The width of the ray in the variable medium can be calculated using the formula:
\( w' = w \cdot \frac{\sin(\theta_1)}{\sin(\theta_2)} \)
Substituting the known values:
\( w' = w \cdot \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2\mu}} = w \cdot \mu \)
Transition to Glass
Next, we need to consider the transition from the variable layer to glass. The refractive index of glass is given as \( \sqrt{2} \). Again applying Snell's Law:
\( \mu \cdot \sin(\theta_2) = \sqrt{2} \cdot \sin(\theta_3) \)
From this, we can find \( \sin(\theta_3) \) and subsequently determine the new width in glass:
\( w'' = w' \cdot \frac{\sin(\theta_2)}{\sin(\theta_3)} \)
Final Calculation
To summarize, the width of the ray as it enters the glass can be expressed as:
- Calculate \( w' = w \cdot \mu \) for the variable layer.
- Use Snell's Law to find \( \sin(\theta_3) \) for the transition to glass.
- Finally, calculate \( w'' \) using the relationship between the angles and the widths.
By following these steps, you can determine how the width of the ray changes as it transitions through the different media. This approach not only illustrates the principles of optics but also emphasizes the importance of understanding how light interacts with materials of varying refractive indices.
