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Chapter 1: Problem 39

A light beam in air is incident on the surface of a pond, making an angle of \(20^{\circ}\) with respect to the surface. What are the angles of reflection and refraction?

Short Answer

Expert verified
The angle of reflection is \(20^{\circ}\) (equal to the angle of incidence), and the angle of refraction is approximately \(14.3^{\circ}\) (calculated using Snell's Law with refractive indices of air and water as 1.0 and 1.33, respectively).

Step by step solution

01

Identify the given information

Given the angle of incidence is \(20^{\circ}\). The refractive indices of air and water are taken as \(n_{air} = 1.0\) (approximately) and \(n_{water} = 1.33\), respectively.
02

Use the law of reflection to find the angle of reflection

The law of reflection states that the angle of incidence is equal to the angle of reflection. Therefore, the angle of reflection \(\theta_{reflection}\) is: \[ \theta_{reflection} = \theta_{incidence} = 20^{\circ} \]
03

Apply Snell's Law to find the angle of refraction

Snell's Law states that for light passing from one medium to another, \[ n_{1} \times \sin\theta_{1} = n_{2} \times \sin\theta_{2} \] Where \(n_{1}\) and \(n_{2}\) are the refractive indices of the mediums, and \(\theta_{1}\) and \(\theta_{2}\) are the angles of incidence and refraction, respectively.
04

Substitute given values into Snell's Law formula

We know the angle of incidence, \(\theta_{1} = 20^{\circ}\), and the refractive indices of air and water, \(n_{1} = 1.0\) and \(n_{2} = 1.33\). We can rewrite Snell's Law as: \[ 1.0 \times \sin(20^{\circ}) = 1.33 \times \sin\theta_{2} \]
05

Solve for the angle of refraction

Now, we will solve for \(\theta_{2}\), the angle of refraction: \[ \theta_{2} = \arcsin \frac{\sin(20^{\circ})}{1.33} \] Calculate the value: \[ \theta_{2} \approx 14.3^{\circ} \]
06

Present the final results

The angle of reflection is \(20^{\circ}\), and the angle of refraction is approximately \(14.3^{\circ}\).

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Most popular questions from this chapter

If you have completely polarized light of intensity \(150 \mathrm{W} / \mathrm{m}^{2},\) what will its intensity be after passing through a polarizing filter with its axis at an \(89.0^{\circ}\) angle to the light's polarization direction?

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