Question

When a polarizing lens is rotated through an angle \(\theta\) over a second lens, the amount of light passing through both lenses decreases by \(1-\sin ^{2} \theta\) a) Determine an equivalent expression for this decrease using only cosine. b) What fraction of light is lost when \(\theta=\frac{\pi}{6} ?\) c) What percent of light is lost when \(\theta=60^{\circ} ?\)

Short answer

a) \( \text{cos}^{2} \theta \) b) \( \frac{1}{4} \) c) 75\%

Step-by-step solution

- Understanding the given expression

The amount of light passing through decreases by an amount described by the expression: \[ 1 - \text{sin}^{2} \theta \]

- Applying Pythagorean identity

Use the Pythagorean trigonometric identity: \[ \text{sin}^{2} \theta + \text{cos}^{2} \theta = 1 \] to substitute for \text{sin}^{2} \theta. \ We can rewrite \text{sin}^{2} \theta \ as \text{1 - cos}^{2} \theta .

- Replace in the original expression

Substitute \ \( 1 - \text{cos}^{2} \theta \) \ for \ \( \text{sin}^{2} \theta \) \ in the equation: \ \[ 1 - (1 - \text{cos}^{2} \theta ) \ \] which simplifies to \ \( \text{cos}^{2} \theta . \)

- Equivalent expression

Thus, the equivalent expression using cosine is \[ \text{cos}^{2} \theta. \]

- Calculate fraction of light lost for \ \( \theta = \frac{\pi}{6} \)

First, find \ \( \text{cos} \frac{\pi}{6} \), which is \ \( \frac{\sqrt{3}}{2}. \) Next, square this value: \ \( \text{cos}^{2} \frac{\pi}{6} = \left( \frac{\sqrt{3}}{2} \right)^{2} = \frac{3}{4}. \) Hence, the fraction of light lost is \ \( 1 - \frac{3}{4} = \frac{1}{4}. \)

- Calculate percent of light lost for \ \( \theta = 60^{\circ} \)

\ \( 60^{\circ} = \frac{\pi}{3} \). Find \ \( \text{cos} 60^{\circ} \), which is \ \( \frac{1}{2}. \) Square this value: \ \( \text{cos}^{2} 60^{\circ} = \left( \frac{1}{2} \right)^{2} = \frac{1}{4}. \) Hence, the fraction of light passed is \ \( \frac{1}{4} \), so the fraction of light lost is \ \( 1 - \frac{1}{4} = \frac{3}{4}. \) Convert this to a percentage: \ \( \frac{3}{4} \times 100 \% = 75 \%. \)

Key concepts

Polarizing Lens

A polarizing lens is a special type of optical lens used to filter light waves. It only allows light waves vibrating in a certain direction to pass through. When you have two polarizing lenses and rotate one of them with respect to the other, the intensity of the light that passes through can change based on the angle \(\theta\) between them. This behavior is governed by trigonometric functions, which describe how the light's properties change with the angle.

Light Intensity

Light intensity measures the amount of light passing through a certain area. When two polarizing lenses are aligned, the light intensity is at its maximum. As one lens is rotated, the intensity reduces based on the sine and cosine of the angle \(\theta\) between the lenses. The equation \(1 - \sin^{2} \theta\) describes how the light intensity decreases with angle.

Trigonometric Functions

Trigonometric functions like sine (\( \sin \)) and cosine (\( \cos \)) are fundamental in understanding wave behavior, including light. In our exercise, the sine function was initially used to describe the decrease in light intensity. Using the Pythagorean identity, \( \sin^{2} \theta + \cos^{2} \theta = 1\), we could convert the expression to use only the cosine function. This way, we find that the light passing through can alternatively be described by \( \cos^{2} \theta \).

Pythagorean Identity

The Pythagorean Identity is a fundamental trigonometric equation: \( \sin^{2} \theta + \cos^{2} \theta = 1 \). This identity is very useful as it allows us to express the \(\text{sin}^{2} \theta\) in terms of \(\text{cos}^{2} \theta\) and vice versa. For example, \(\text{sin}^{2} \theta\) can be written as \(1 - \cos^{2} \theta\). This conversion is particularly helpful in problems where one trigonometric function is easier to calculate or understand.