Chapter 37: Problem 4
The refractive index of diamond is \(2.42 .\) What is the critical angle for light passing from diamond to air? We use \(n_{i} \sin \theta_{i}=n_{t} \sin \theta_{t}\) to obtain $$ (2.42) \sin \theta_{c}=(1) \sin 90^{\circ} $$ from which it follows that \(\sin \theta_{c}=0.413\) and \(\theta_{c}=24.4^{\circ}\).
Short Answer
Step by step solution
Identify the Known Values
Understand the Critical Angle Formula
Rearrange the Equation
Calculate the Critical Angle
Determine \( \theta_c \) using Inverse Sine Function
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Refractive Index
- Air has a refractive index close to 1.
- Water has a refractive index of about 1.33.
- Diamond has a notably high refractive index of 2.42.
Snell's Law
- \( n_i \) and \( n_t \) are the refractive indexes of the initial and target media, respectively.
- \( \theta_i \) is the angle of incidence, the angle between the incoming light and the normal to the boundary.
- \( \theta_t \) is the angle of refraction, the angle between the refracted light and the normal.
Optics
- Reflection occurs when light bounces off a surface.
- Refraction is the bending of light as it passes through different media. It is essential in lenses, prisms, and various optical devices.
- Interference refers to the overlap of different light waves, leading to patterns of dark and bright fringes.
- Diffraction is the spreading out of light waves when they encounter obstacles or slits.