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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

Suppose your height is \(2.0 \mathrm{~m}\) and you are standing \(50 \mathrm{~cm}\) in front of a plane mirror. a) What is the image distance? b) What is the image height? c) Is the image inverted or upright? d) Is the image real or virtual?

Standing by a pool filled with water, under what condition will you see a reflection of the scenery on the opposite side through total internal reflection of the light from the scenery? a) Your eyes are level with the water. b) You observe the pool at an angle of \(41.8^{\circ}\) c) Under no condition. d) You observe the pool at an angle of \(48.2^{\circ}\)

A \(5.00-\mathrm{cm}\) object is placed \(30.0 \mathrm{~cm}\) away from a convex mirror with a focal length of \(-10.0 \mathrm{~cm}\). Determine the size, orientation, and position of the image.

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An optical fiber with an index of refraction of 1.5 is used to transport light of wavelength \(400 \mathrm{nm}\). What is the critical angle for light to transport through this fiber without loss? If the fiber is immersed in water? In oil?
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