Problem 41
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?
Step-by-Step Solution
Verified Answer
The maximum possible index of refraction of the oil is approximately 1.28.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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: \(n_{glass} \sin(\theta_c) = n_{oil}\cdot 1\).
4Step 4: 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 \(\theta_c\).
5Step 5: 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: \(n_{oil} = 1.52 \times \sin(57.2°)\).
6Step 6: 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 LawCritical AngleIndex of Refraction
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.
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.
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.
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