Problem

# In each of the following cases, deduce the nature of the light that is consistent with the...

In each of the following cases, deduce the nature of the light that is consistent with the analysis performed. Assume a 100% efficient polarizer.

a. When a polarizer is rotated in the path of the light, there is no intensity variation. With a QWP in front of the rotating polarizer (coming first), one finds a variation in intensity but no angular position of the polarizer that gives zero intensity.

b. When a polarizer is rotated in the path of the light, there is some intensity variation but no position of the polarizer giving zero intensity. The polarizer is set to give maximum intensity. A QWP is allowed to intercept the beam first with its OA parallel to the TA of the polarizer. Rotation of the polarizer now can produce zero intensity.

#### Step-by-Step Solution

Solution 1

(a)

When an unpolarised light of irradiance $$I_{0}$$ is incident on a polarizer, the irradiance of transmitted ray is $$\frac{1}{2} I_{0}$$ along $$\mathrm{TA}_{1}$$ and irradiance absorbed will be $$\frac{1}{2} I_{0}$$ along perpendicular to $$\mathrm{TA}_{1}$$.

Here given that the transmitted irradiance is $$\alpha$$ times the original one.

Therefore the irradiance of transmitted ray is $$\frac{1}{2} \alpha I_{0}$$ along $$\mathrm{TA}_{1}$$.

Also $$\beta$$ part of absorbed irradiance is transmitted along perpendicular to $$\mathrm{TA}_{1}$$.

Transmitted irradiance along perpendicular to $$\mathrm{TA}_{1}$$ is $$\frac{1}{2} \beta I_{0}$$.

Now the two polarized lights are incident on second analyzer which is at an angle $$\theta$$ with first polarizer.

The two polarized light irradiations are

1. $$\frac{1}{2} \alpha I_{0}$$ Incident at an angle $$\theta$$.

2. $$\frac{1}{2} \beta I_{0}$$ Incident at an angle $$90-\theta$$.

(1)

For the irradiation incident at an angle $$\theta$$ with $$\mathrm{TA}_{2}$$ there are two transmitted irradiations are

(i) $$\alpha\left(\frac{1}{2} \alpha I_{0}\right) \cos ^{2} \theta=\frac{1}{2} \alpha^{2} I_{0} \cos ^{2} \theta$$ and

(ii) $$\beta\left[\frac{1}{2} \alpha I_{0}-\frac{1}{2} \alpha I_{0} \cos ^{2} \theta\right]=\frac{1}{2} \beta \alpha I_{0}\left(1-\cos ^{2} \theta\right)$$ $$=\frac{1}{2} \alpha \beta I_{0} \sin ^{2} \theta$$

Therefore total transmitted irradiance due to the incident irradiance $$\frac{1}{2} \alpha I_{0}$$ is $$\frac{1}{2} \alpha^{2} I_{0} \cos ^{2} \theta+\frac{1}{2} \alpha \beta I_{0} \sin ^{2} \theta$$

(2)

For the irradiation incident at an angle $$90-\theta$$ with $$\mathrm{TA}_{2}$$ there are two transmitted irradiations.

(i) $$\alpha\left(\frac{1}{2} \beta I_{0} \cos ^{2}(90-\theta)\right)=\frac{1}{2} \alpha \beta I_{0} \sin ^{2} \theta$$ and

(ii) $$\beta\left[\frac{1}{2} \beta I_{0}-\frac{1}{2} \beta \cos ^{2}(90-)\right]=\frac{1}{2} \beta^{2} I_{0}\left[1-\sin ^{2} \theta\right]$$

$$=\frac{1}{2} \beta^{2} I_{0} \cos ^{2} \theta$$

Therefore the total transmitted irradiance due to the incident irradiation $$\frac{1}{2} \beta I_{0}$$ is $$\frac{1}{2} \alpha \beta I_{0} \sin ^{2} \theta+\frac{1}{2} \beta^{2} I_{0} \cos ^{2} \theta$$

\begin{aligned} I &=\frac{1}{2} \alpha^{2} I_{0} \cos ^{2} \theta+\frac{1}{2} \alpha \beta I_{0} \sin ^{2} \theta+\frac{1}{2} \alpha \beta I_{0} \sin ^{2} \theta+\frac{1}{2} \beta^{2} I_{0} \cos ^{2} \theta \\ &=\frac{1}{2} I_{0}\left(\alpha^{2}+\beta^{2}\right) \cos ^{2} \theta+\alpha \beta I_{0} \sin ^{2} \theta \\ &=I_{0}\left[\frac{1}{2}\left(\alpha^{2}+\beta^{2}\right) \cos ^{2} \theta+\alpha \beta \sin ^{2} \theta\right] \end{aligned}

Therefore

$$I=I_{0}\left[0.5\left(\alpha^{2}+\beta^{2}\right) \cos ^{2} \theta+\alpha \beta \sin ^{2} \theta\right]$$

For ideal case $$\alpha=1$$ and $$\beta=0$$

Therefore

\begin{aligned} I &=I_{0}\left[0.5\left(1^{2}+0^{2}\right) \cos ^{2} \theta+(1) \sin ^{2} \theta\right] \\ &=0.5 I_{0} \cos ^{2} \theta \ldots \text { (1) } \end{aligned}

In the ideal case when unpolarised light of irradiation $$I_{0}$$ incident on a polarizer the intensity of transmitted irradiance is $$\frac{1}{2} I_{0}$$ and when this transmitted irradiance is incident on a polarizer which is at an angle $$\theta$$ with the first polarizer, then transmitted irradiance will be

$$I=\left(\frac{1}{2} I_{0}\right) \cos ^{2} \theta$$

Therefore, $$I=0.5 I_{0} \cos ^{2} \theta \cdots \cdots$$ (2)

1 and 2 are identical.

(b)

Given $$\alpha=0.95$$ and $$\beta=0.05$$

Then the transmitted irradiance will be

\begin{aligned} I &=I_{0}\left[(0.5)\left[\left(0.95^{2}+0.0025\right)\right] \cos ^{2} \theta+0.0475 \sin ^{2} \theta\right] \\ &=I_{0}\left[(0.5)(0.9025+0.0025) \cos ^{2} \theta+0.0475 \sin ^{2} \theta\right] \\ &=I_{0}\left[(0.4525) \cos ^{2} \theta+0.0475 \sin ^{2} \theta\right] \end{aligned}

And for ideal case $$I=0.5 I_{0} \cos ^{2} \theta$$

\text { (i) } \begin{aligned} \theta &=0^{\circ} \\ I &=I_{0}\left[(0.4525)\left(\cos ^{2} \theta\right)+(0.0475)\left(\sin ^{2} \theta\right)\right] \\ =& I_{0}[(0.4525)(1)+(0.04725)(0)] \\ =& 0.4525 I_{0} \\ I_{\text {ideal }} &=0.5 I_{0} \cos ^{2}(0) \\ &=0.5 I_{0} \end{aligned}

\text { (ii) } \begin{aligned} \theta &=30^{\circ} \\ I &=I_{0}\left[(0.4525)\left(\cos ^{2} \theta\right)+(0.0475)\left(\sin ^{2} \theta\right)\right] \\ &=I_{0}\left[(0.4525)\left(\cos ^{2} 30\right)+(0.0475)\left(\sin ^{2} 30\right)\right] \\ &=I_{0}\left[(0.4525)(0.866)^{2}+(0.04725)(0.5)^{2}\right] \\ &=0.3512 I_{0} \\ I_{\text {ideal }} &=0.5 I_{0} \cos ^{2}(30) \\ &=0.375 I_{0} \end{aligned}

\text { (iii) } \begin{aligned} \theta &=45^{\circ} \\ I &=I_{0}\left[(0.4525)\left(\cos ^{2} \theta\right)+(0.0475)\left(\sin ^{2} \theta\right)\right] \\ =& I_{0}\left[(0.4525)\left(\cos ^{2} 45\right)+(0.0475)\left(\sin ^{2} 45\right)\right] \\ =& I_{0}\left[(0.4525)\left(\frac{1}{\sqrt{2}}\right)^{2}+(0.04725)\left(\frac{1}{\sqrt{2}}\right)^{2}\right] \\ =& I_{0}[(0.4525)(0.5)+(0.04725)(0.5)] \\ =& 0.249875 I_{0} \\ I_{\text {ideal }}=& 0.5 I_{0} \cos ^{2}(45) \\ =& 0.25 I_{0} \end{aligned}

(iv) $$\theta=90^{\circ}$$

\begin{aligned} I &=I_{0}\left[(0.4525)\left(\cos ^{2} \theta\right)+(0.0475)\left(\sin ^{2} \theta\right)\right] \\ &=I_{0}\left[(0.4525)\left(\cos ^{2} 90\right)+(0.0475)\left(\sin ^{2} 90\right)\right] \\ &=I_{0}\left[(0.4525)(0)^{2}+(0.04725)(1)^{2}\right] \\ &=0.04725 I_{0} \\ I_{\text {ideal }} &=0.5 I_{0} \cos ^{2}(90) \\ &=0 \end{aligned}

Solutions For Problems in Chapter 15
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