Question

Which of the following interface combinations has the smallest critical angle? a) light traveling from ice to diamond b) light traveling from quartz to lucite c) light traveling from diamond to glass d) light traveling from lucite to diamond e) light traveling from lucite to quartz

Short answer

Answer: The smallest critical angle is found in interface combination c) light traveling from diamond to glass, with a critical angle of approximately 38.60°.

Step-by-step solution

List the refractive indices of all materials

Look up and write down the refractive indices of all the materials involved in the interfaces - ice, diamond, quartz, lucite, and glass: - Refractive index of ice: n_ice = 1.31 - Refractive index of diamond: n_diamond = 2.42 - Refractive index of quartz: n_quartz = 1.54 - Refractive index of lucite: n_lucite = 1.50 - Refractive index of glass: n_glass = 1.52

Calculate the critical angle for each interface combination

For each interface, apply the formula 𝜃c = arcsin(n2/n1) to determine the critical angle: a) Ice to diamond: 𝜃c = arcsin(n_diamond/n_ice) = arcsin(2.42 / 1.31) (This combination, as the refractive indices are wrongly swapped, is not possible.) b) Quartz to lucite: 𝜃_c = arcsin(n_lucite/n_quartz) = arcsin(1.50 / 1.54) c) Diamond to glass: 𝜃_c = arcsin(n_glass/n_diamond) = arcsin(1.52 / 2.42) d) Lucite to diamond: 𝜃_c = arcsin(n_diamond/n_lucite) = arcsin(2.42 / 1.50) (This combination, as the refractive indices are wrongly swapped, is not possible.) e) Lucite to quartz: 𝜃_c = arcsin(n_quartz/n_lucite) = arcsin(1.54 / 1.50)

Compare critical angles to determine the smallest

Now we determine which of the valid combinations b), c), and e) has the smallest critical angle by calculating the angles: b) 𝜃_c = arcsin(1.50 / 1.54) ≈ 78.46° c) 𝜃_c = arcsin(1.52 / 2.42) ≈ 38.60° e) 𝜃_c = arcsin(1.54 / 1.50) ≈ 80.22° The smallest critical angle is found in interface combination c) light traveling from diamond to glass, with a critical angle of approximately 38.60°.

Key concepts

Refractive Index

Understanding the refractive index is crucial in the study of optics. It is a dimensionless number that describes how light propagates through a medium compared to the speed of light in a vacuum. In essence, it measures the bending, or refraction, of light when it travels from one medium to another.

The refractive index (denoted as 'n') quantifies this effect and depends on the optical density of the material and the wavelength of the light entering. For any medium, the refractive index is calculated as the ratio of the speed of light in a vacuum (approximately 299,792 kilometers per second) to the speed of light in the material. Mathematically, this is expressed as:
$$ n = \frac{c}{v} $$ where 'c' is the speed of light in a vacuum, and 'v' is the speed of light in the material.

In the context of the textbook problem provided, different materials have varying refractive indices, which affect the critical angle when light moves between them. For instance, diamond has a high refractive index, indicating that light travels through it more slowly and bends more when exiting into materials with lower refractive indices, like air or glass.

Total Internal Reflection

Total internal reflection is a fascinating optical phenomenon that occurs when a light ray travels from a medium with a higher refractive index to one with a lower refractive index and hits the boundary at an angle greater than the so-called critical angle. This causes the light to reflect completely back into the original medium instead of being refracted or passing through into the second medium.

This is what makes fiber optic cables function so effectively; they lead light over long distances with minimal signal loss. The principle is also employed in binoculars, periscopes, and some types of prisms.

Conditions for Total Internal Reflection

• The light must travel from a more optically dense medium to a less optically dense medium.
• The angle of incidence must be greater than the critical angle for the specific material boundary.

The exercise provided asks for the smallest critical angle among different material interfaces. The importance of the critical angle is evident because it determines the threshold at which this total internal reflection occurs.

Snell's Law

Snell's Law, named after the Dutch mathematician Willebrord Snellius, is a formula used to describe the relationship between the angles of incidence and refraction for a wave (e.g., light) passing through a boundary between two different isotropic media. It encapsulates how light bends at the interface of two materials with different refractive indices and can be written as:
$$ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) $$where:

• \(n_1\) is the refractive index of the first medium,
• \(\theta_1\) is the angle of incidence,
• \(n_2\) is the refractive index of the second medium, and
• \(\theta_2\) is the angle of refraction.

When there's no refraction at the boundary and the light is fully reflected, the angle of incidence is the critical angle. Applying Snell's Law at this critical angle and knowing that the angle of refraction would be 90 degrees, we can derive the formula for the critical angle as:$$ \theta_c = \arcsin(\frac{n_2}{n_1}) $$In the provided solution, this relationship was used to calculate the critical angles for different material combinations. Accurately understanding and applying Snell's Law is vital for solving a wide range of optical problems.