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Chapter 18: Problem 150

Unpolarized light of intensity \(32 \mathrm{Wm}^{-2}\) passes through three polarizers such that transmission axes of the first and second polarizer makes an angle \(30^{\circ}\) with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of final emerging light will be

Short Answer

Expert verified
The intensity of the final emerging light after passing through three polarizers will be \(0 \mathrm{Wm}^{-2}\)

Step by step solution

01

Intensity after first polarizer

The intensity of the light after passing through the first polarizer is half of the initial intensity, according to the properties of an ideal polarizer. Hence, it would be \(32 \mathrm{Wm}^{-2} / 2 = 16 \mathrm{Wm}^{-2}\)
02

Intensity after second polarizer

We apply Malus' Law, which gives the intensity after the light passes through the second polarizer as \(I_{2} = I_{1} \cdot cos^{2}(\theta)\). Here, \(I_{1}\) is the intensity after the first polarizer and \( \theta \) is the angle between the first and second polarizer, 30 degrees. Hence, \( I_{2} = 16 \mathrm{Wm}^{-2} \cdot cos^{2}(30^{\circ}) = 16 \mathrm{Wm}^{-2} \cdot (\sqrt{3}/2)^{2} = 12 \mathrm{Wm}^{-2} \)
03

Intensity after third polarizer

Now, we apply Malus' Law to the light pass through the third polarizer. The transmission axis of the third polarizer is crossed with respect to first (or at \(90^{\circ}\). Therefore, the final intensity \(I_{3}\) will be \( I_{2} \cdot cos^{2}(90^{\circ}) = 12 \mathrm{Wm}^{-2} \cdot (0)^{2} = 0 \mathrm{Wm}^{-2}\)

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Most popular questions from this chapter

In a Young's double slit experiment, the intensity at a point where the path difference is \(\frac{\lambda}{6}(\lambda\) being the wavelength of light used) is \(I\). If \(I_{0}\) denotes the maximum intensity, \(\frac{I}{I_{0}}\) is equal to [2007] (A) \(\frac{3}{4}\) (B) \(\frac{1}{\sqrt{2}}\) (C) \(\frac{\sqrt{3}}{2}\) (D) \(\frac{1}{2}\)

To get three images of single object, one should have two plane mirrors at an angle of (A) \(60^{\circ}\) (B) \(90^{\circ}\) (C) \(120^{\circ}\) (D) \(30^{\circ}\)

\(x\)-axis is normal to the reflecting surface of a plane mirror. If object is moving with velocity \((3 \hat{i}+4 \hat{k}) \mathrm{m} / \mathrm{s}\) The relative velocity of image with respect to object will be along (A) \(-x\) axis (B) \(+x\) axis (C) \(-z\) axis (D) \(+z\) axis

In the diagram shown, the object is performing SHM according to the equation \(y=2 A \sin (\omega t)\) and the plane mirror is performing SHM according to the equation \(Y=-A \sin \left(\omega t-\frac{\pi}{3}\right)\). The diagram shows the state of the object and the mirror at time \(t=0 \mathrm{~s}\). The minimum time from \(t=0 \mathrm{~s}\) after which the velocity of the image becomes equal to zero is (A) \(\frac{\pi}{3 \omega}\) (B) \(\frac{3 \pi}{\omega}\) (C) \(\frac{\pi}{6 \omega}\) (D) \(\frac{2 \pi}{3 \omega}\)

In an astronomical telescope, the focal length of objective lens and eyepiece are \(150 \mathrm{~cm}\) and \(6 \mathrm{~cm}\), respectively. In case when final image is formed at least distance of clear vision \((D=25 \mathrm{~cm})\). The magnifying power is (A) 29 (B) 30 (C) 31 (D) 32
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