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Chapter 23: Problem 35

A defect in a diamond appears to be 2.0 mm below the surface when viewed from directly above that surface. How far beneath the surface is the defect?

Short Answer

Expert verified
Answer: The actual depth of the defect in the diamond is approximately 4.84 mm.

Step by step solution

01

Identify the given information

The apparent depth of the defect in the diamond is 2.0 mm. Let's denote this depth as \(d_{apparent}\). The index of refraction for air (the medium from which we are looking into the diamond) is approximately 1.00, and the index of refraction for diamond is approximately 2.42. We'll denote these values as \(n_{air}\) and \(n_{diamond}\), respectively.
02

Apply Snell's Law to the system

According to Snell's Law, the following equality holds: $$n_{air} \cdot \sin (\theta_i) = n_{diamond} \cdot \sin (\theta_r)$$ In this case, the angle of incidence \(\theta_i\) is 90 degrees because we are viewing directly above the surface, so \(\sin (\theta_i) = 1\). We can rewrite Snell's Law as follows: $$1.00 \cdot \sin (90^\circ) = 2.42 \cdot \sin (\theta_r)$$ Now, we can solve for \(\theta_r\): $$\sin (\theta_r) = \frac{1}{2.42}$$ $$\theta_r = \sin^{-1} \left(\frac{1}{2.42}\right)$$
03

Use the refracted angle to find the actual depth

We will now use some trigonometry to find the actual depth, denoted as \(d_{actual}\). Looking at the problem as a right triangle, we can see that the ratio between the actual depth of the defect and the apparent depth to the surface is equal to the tangent of the angle of refraction (\(\theta_r\)): $$\tan (\theta_r) = \frac{d_{actual}}{d_{apparent}}$$ Now, we can solve for the actual depth \(d_{actual}\): $$d_{actual} = d_{apparent} \cdot \tan (\theta_r) = 2.0 \text{ mm} \cdot \tan \left(\sin^{-1} \left(\frac{1}{2.42}\right)\right)$$ Calculating this value, we find that the defect is approximately 4.84 mm beneath the surface.

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