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Chapter 32: Problem 49

One of the factors that cause a diamond to sparkle is its relatively small critical angle. Compare the critical angle of diamond in air compared to that of diamond in water.

Short Answer

Expert verified
Answer: The critical angle for a diamond in air is approximately 24.4°, whereas the critical angle for a diamond in water is approximately 32.8°. The diamond in air has a smaller critical angle, which means it experiences more total internal reflection, causing it to sparkle more.

Step by step solution

01

First, we need to identify the refractive indices of the materials involved. The refractive index of a diamond is approximately 2.42. The refractive index of air is close to 1, and the refractive index of water is approximately 1.33. #Step 2: Calculate the Critical Angle: Diamond in Air#

The critical angle (theta_c) is the angle of incidence for which the angle of refraction is 90 degrees. To find the critical angle, we can use Snell's Law: \(n_1 \cdot \sin(\theta_c) = n_2 \cdot \sin(90^{\circ})\) Where \(n_1\) is the refractive index of the first medium (diamond), \(n_2\) is the refractive index of the second medium(air), and \(\theta_c\) is the critical angle. In our case, \(n_1 = 2.42\) and \(n_2 = 1\). Solving for \(\theta_c\), we get: \(\theta_c = \arcsin(\frac{n_2}{n_1}) = \arcsin(\frac{1}{2.42})\) #Step 3: Calculate the Critical Angle: Diamond in Water#
02

Now we will calculate the critical angle for a diamond in water. We will use the same formula and replace \(n_2\) with the refractive index of water, which is 1.33. So, we have \(n_1 = 2.42\) and \(n_2 = 1.33\). Solving for \(\theta_c\): \(\theta_c = \arcsin(\frac{n_2}{n_1}) = \arcsin(\frac{1.33}{2.42})\) #Step 4: Compare the Critical Angles#

We can now compare the critical angles for a diamond in air and a diamond in water: Diamond in air: \(\theta_c = \arcsin(\frac{1}{2.42}) \approx 24.4^{\circ}\) Diamond in water: \(\theta_c = \arcsin(\frac{1.33}{2.42}) \approx 32.8^{\circ}\) The critical angle for a diamond in air is smaller than the critical angle for a diamond in water. This means that a diamond will experience more total internal reflection in air, making it sparkle more.

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Most popular questions from this chapter

A \(45^{\circ}-45^{\circ}-90^{\circ}\) triangular prism can be used to reverse a light beam: The light enters perpendicular to the hypotenuse of the prism, reflects off each leg, and emerges perpendicular to the hypotenuse again. The surfaces of the prism are not silvered. If the prism is made of glass with in dex of refraction \(n_{\text {glass }}=1.520\) and the prism is surrounded by air, the light beam will be reflected with a minimum loss of intensity (there are reflection losses as the light enters and leaves the prism). a) Will this work if the prism is under water, which has index of refraction \(n_{\mathrm{H}_{2} \mathrm{O}}=1.333 ?\) b) Such prisms are used, in preference to mirrors, to bend the optical path in quality binoculars. Why?

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