Chapter 38, Problem 48P

(a)

To determine

The fraction by which the transmitted intensity is reduced.

Answer to Problem 48P

The intensity will be reduced by $0.875$ fraction of the incident intensity.

Explanation of Solution

Given info: The number of the polarizing filters is $3$. The formula to calculate the intensity after it passes through any polarizing filter is,

$I=I_0{\cos }^2\left(\theta \right)$ . (1)

Here,

$I$ is the intensity that came out of the polarizing filter.

$I_0$ is the incident intensity of the un polarized light.

$\theta$ is the angle between the transmission of the two polarizing filters.

When an unpolarized light is passed through a polarizing filter intensity is reduced to half. So after passing through the first polarizer the intensity of the light becomes half.

$I_{01}=\frac{I_0}{2}$

Here,

$I_{01}$ is the intensity of the light after the first polarizing filter

The angle between the transmission axis of second polarizer and the first polarizer is $45.0°$. Therefore, from equation (1) the formula to calculate the intensity when the light comes out of the second polarizer is,

$I_{02}=I_{01}{\left(\cos \theta \right)}^2$ (2)

Here,

$I_{02}$ is the intensity of the light after the second  polarizing filter

The third polarizing filter and the second polarizing filter has the same $45.0°$ angle in between their transmission axis is also

Therefore the final intensity after three polarizing filters is,

$I_{03}=I_{02}{\left(\cos \theta \right)}^2$ (3)

Substitute $I_{01}{\left(\cos \theta \right)}^2$ for $I_{02}$ in equation (3),

$\begin{array}{l}{I}_{03}=I_{02}{\left(\cos \theta \right)}^2\\ I_{03}=I_{01}{\left(\cos \theta \right)}^2{\left(\cos \theta \right)}^2\end{array}$ (4)

Substitute $\frac{I_0}{2}$ for $I_{01}$ in equation (4),

$\begin{array}{l}{I}_{03}=I_{01}{\left(\cos \theta \right)}^2{\left(\cos \theta \right)}^2\\ I_{03}=\frac{I_0}{2}{\left(\cos \theta \right)}^2{\left(\cos \theta \right)}^2\end{array}$ (5)

From equation (5), a general formula for the calculation of intensity when light is passed through $n$ number of polarizing filters is,

$I=\frac{I_0}{2}{\left({\cos }^2\theta \right)}^{n-1}$ (6)

Here,

$I$ is the final intensity.

$I_0$ is the initial intensity.

$n$ is the number of polarizing filters.

Substitute $45.0°$ for $\theta$ in equation 5 to calculate the fraction of intensity transmitted after three polarizing filters,

$\begin{array}{c}{I}_{03}=\frac{I_0}{2}{\left(\cos \theta \right)}^2{\left(\cos \theta \right)}^2\\ =\frac{I_0}{2}{\left(\cos 45°\right)}^2\left(\cos 45°\right)\\ =\frac{I_0}{2}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\\ =0.125I_0\end{array}$

Therefore the absorbed intensity is

$\begin{array}{c}{I}_{\text{abs}}=I_0-0.125I_o\\ =0.875I_0\end{array}$ (7)

Conclusion:

Therefore, the fraction by which the intensity is reduced is $0.875$ of the incident intensity.

(b)

To determine

The fraction by which intensity is reduced when $4$ polarizing filters are placed.

Answer to Problem 48P

The fraction by which the intensity is reduced is $0.789$ of the incident intensity.

Explanation of Solution

Given info: The number of filters are $4$ and the angle between the transmission axis is $30.0°$.

From equation (6) the formula to calculate when there are $n$ number of polarizing filters are present,

$I=\frac{I_0}{2}{\left({\cos }^2\theta \right)}^{n-1}$

Substitute $30.0°$ for $\theta$ and $4$ for $n$ in the above equation,

$\begin{array}{l}I=\frac{I_0}{2}{\left({\cos }^2\theta \right)}^{n-1}\\ I=\frac{I_{02}}{}{\left({\cos }^{2}30.0°\right)}^{4-1}\\ I=I_0\left(0.211\right)\end{array}$

Therefore the absorbed intensity is

$\begin{array}{c}{I}_{\text{abs}}=I_0-0.211I_0\\ =0.789I_0\end{array}$ (8)

Conclusion:

Therefore, The fraction by which the intensity is reduced is $0.789$ of the incident intensity

(c)

To determine

The fraction by which intensity is reduced when $7$ polarizing filters are placed.

Answer to Problem 48P

The fraction by which the intensity is reduced is $0.670$ of the incident intensity.

Explanation of Solution

Given info: The number of filters are $7$ and the angle between the transmission axis is $15.0°$.

From equation (6) the formula to calculate when there are $n$ number of polarizing filters are present,

$I=\frac{I_0}{2}{\left({\cos }^2\theta \right)}^{n-1}$

Substitute $15.0°$ for $\theta$ and $7$ for $n$ in the above equation,

$\begin{array}{l}I=\frac{I_0}{2}{\left({\cos }^2\theta \right)}^{n-1}\\ I=\frac{I_{02}}{}{\left({\cos }^{2}15.0°\right)}^{7-1}\\ I=I_0\left(0.330\right)\end{array}$

Therefore the absorbed intensity is

$\begin{array}{c}{I}_{\text{abs}}=I_0-0.330I_0\\ =0.670I_0\end{array}$ (9)

Conclusion:

Therefore, the fraction by which the intensity is reduced is $0.670$ of the incident intensity

(d)

To determine

The comparison between answer of part (a), (b) and (c).

Answer to Problem 48P

The intensity of light can be increased by increasing the number of stacks of polarizing filters by decreasing the angle between their transmission axis.

Explanation of Solution

From equation (7), (8) and (9), it is evident that, as the number of polarizing filters increased the fraction of absorbed was decreased. For the case of $3$ polarizing filters and angle between the transmission axis $45.0°$ the absorbed intensity was $0.875$ of the incident intensity, for the case of $4$ filters and angle between the transmission axis $30.0°$ the absorption was reduced to $0.789$ of the incident intensity and finally for the case of $7$ filters and angle between the transmission axis $15.0°$ the absorption was only $0.670$ of the incident intensity.

Conclusion:

Therefore, the intensity of light can be increased by increasing the number of stacks of polarizing filters by decreasing the angle between their transmission axis.