Question

A ray of light traveling \(in\) a block of glass (\(n\) = 1.52) is incident on the top surface at an angle of 57.2\(^\circ\) with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

Short answer

The maximum possible index of refraction of the oil is approximately 1.28.

Step-by-step solution

Identify the Scenario

We have a ray of light traveling through a block of glass with a refractive index of 1.52. This ray hits the glass-oil interface at an angle of 57.2° with respect to the normal.

Understanding Total Internal Reflection

For total internal reflection to occur at the interface between glass and oil, the critical angle must be reached. The critical angle is where the light goes from glass to oil and refracts at 90°, thus reflecting back into the glass.

Apply Snell's Law

Snell's Law is given by \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\). At the critical angle \(\theta_c\), we have \(\sin(\theta_2) = 1\), since it's refracting along the interface. The equation at the critical angle becomes: \(n_{glass} \sin(\theta_c) = n_{oil}\cdot 1\).

Calculate the Critical Angle

Since the oil surface causes total internal reflection, find the critical angle from glass to oil using: \(\theta_c = \sin^{-1}(\frac{n_{oil}}{n_{glass}})\). With total internal reflection happening at 57.2°, that's our \(\theta_c\).

Solve for the Index of Refraction of Oil

Rearrange the formula for \(n_{oil}\) to find the maximum possible index. Since \(\theta_c = 57.2°\), set \(\sin(\theta_c) = \frac{n_{oil}}{1.52}\). Substituting for \(\theta_c\), solve: \(n_{oil} = 1.52 \times \sin(57.2°)\).

Calculate Sin(57.2°) and Determine \(n_{oil}\)

Calculate \(\sin(57.2°)\) and multiply by the refractive index of the glass 1.52 to find \(n_{oil}\). \(\sin(57.2°)\approx 0.842\). Calculate: \(n_{oil} \approx 1.52 \times 0.842\approx 1.28\).

Key concepts

Snell's Law

Snell's Law is essential for understanding how light behaves when it transitions between different media. This law is expressed through the equation: \[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]where:

• \(n_1\) and \(n_2\) are the indices of refraction for the first and second media, respectively.
• \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, measured from the normal to the surface.

Snell's Law helps explain why light bends when entering a new medium. The bending occurs due to the change in the speed of light as it moves through materials with different refractive indices. This principle is key in optics and helps design lenses and other optical devices. In the context of our light ray problem, it allows us to determine how light behaves at the glass-oil interface.

Critical Angle

The critical angle is the angle of incidence in the denser medium at which light refracts along the boundary with the less dense medium, rather than passing through. It is calculated using the formula:\[\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)\]This phenomenon only occurs when:

• Light moves from a medium with a higher index of refraction to one of lower index.
• The angle of incidence exceeds this critical threshold.

Once the critical angle is surpassed, total internal reflection occurs, causing the light to be completely reflected back into the original medium. This effect is used in devices like fiber-optic cables and binoculars.In our scenario, with the light traveling from glass (= 1.52) to oil, finding the critical angle helps us determine the conditions necessary for total internal reflection to happen. Since the ray is incident at 57.2° and reflects back, this tells us the critical angle equals 57.2°, limiting the index of refraction of potential oils.

Index of Refraction

The index of refraction, also called the refractive index, measures how much a light beam bends when it enters a medium. It’s denoted by \(n\) and varies with different materials. The formula for the index of refraction is:\[n = \frac{c}{v}\]where:

• \(c\) is the speed of light in a vacuum.
• \(v\) is the speed of light in the medium.

Materials with higher refractive indices slow down light more, causing more bending.This property is foundational in optics, determining how media like glass and water bend light and how lenses converge or diverge rays.In our problem, knowing the refractive index of glass (1.52) allows for calculating the maximum refractive index the oil layer can have, ensuring that total internal reflection still occurs. The formula used here is derived from considering the critical angle, leading to a maximum oil refractive index of roughly 1.28, preventing light from refracting into the oil.