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Chapter 1: Problem 96

Prove that if \(I\) is the intensity of light transmitted by two polarizing filters with axes at an angle \(\theta\) and \(I^{\prime}\) is the intensity when the axes are at an angle \(90.0^{\circ}-\theta\) then \(I+I^{\prime}=I_{0}, \quad\) the original intensity. (Hint: Use the trigonometric identities \(\cos 90.0^{\circ}-\theta=\sin \theta \quad\) and \(\left.\cos ^{2} \theta+\sin ^{2} \theta=1 .\right)\)

Short Answer

Expert verified
We are given that the intensity of light transmitted by two polarizing filters with axes at an angle \( \theta \) is \(I\) and at an angle \(90.0^{\circ}-\theta\) is \(I'\). Using Malus' Law and the trigonometric identity \( \cos(90^\circ - \theta) = \sin \theta \), we get \( I = I_0 \cos^2{\theta} \) and \( I' = I_0 \sin^2{\theta} \). Adding the intensities, we obtain \( I + I' = I_0 \cos^2{\theta} + I_0 \sin^2{\theta} \). Finally, using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we conclude that the intensities of light transmitted by the two polarizing filters add up to the original intensity: \( I + I' = I_0 \).

Step by step solution

01

Write down the intensities

Given that the intensity of light transmitted by two polarizing filters with axes at an angle \( \theta \) is \(I\) and at an angle \(90.0^{\circ}-\theta\) is \(I'\). Using Malus' Law, we have: \( I = I_0 \cos^2{\theta} \) \( I' = I_0 \cos^2{(90^\circ - \theta)} \)
02

Use the trigonometric identity

Now, we'll use the given trigonometric identity \( \cos(90^\circ - \theta) = \sin \theta \), so we have: \( I' = I_0 \cos^2{(90^\circ - \theta)} = I_0 \sin^2{\theta} \)
03

Add the intensities

Next, we'll add the intensities of light transmitted by the two polarizing filters: \( I + I' = I_0 \cos^2{\theta} + I_0 \sin^2{\theta} \)
04

Use the trigonometric identity

We'll now use the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \) , so we have: \( I + I' = I_0 (\cos^2{\theta} + \sin^2{\theta}) \) \( I + I' = I_0 \cdot 1 \)
05

Conclusion

Therefore, the intensities of light transmitted by two polarizing filters do indeed add up to the original intensity: \( I + I' = I_0 \) This completes the proof.

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Most popular questions from this chapter

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