Chapter 24, Problem 61P

Light with a wavelength in vacuum of 546.1 nm falls perpendicularly on a biological specimen that is 1.000 μm thick. The light splits into two beams polarized at right angles, for which the indices of refraction are 1.320 and 1.333, respectively. (a) Calculate the wavelength of each component of the light while it is traversing the specimen. (b) Calculate the phase difference between the two beams when they emerge from the specimen.

To determine

The wavelength of each components of the light while it is traversing the specimen.

Answer to Problem 61P

The first component of light has a wavelength of $413.7\text{nm}$ and the second component of light has a wavelength of $409.7\text{nm}$.

Explanation of Solution

Given Info: The wavelength of incoming light is $546.1\text{nm}$, the refractive index for first beam is $1.320$, and the refractive index for second beam is $1.333$.

Formula to calculate the wavelength of the first beam is,

${\lambda }_{n_1}=\frac{\lambda }{n_1}$

• ${\lambda }_{n_1}$ is the wavelength of the first component
• $\lambda$ is the wavelength of the incident light
• $n_1$ is the refractive index for the first beam

Substitute $546.1\text{nm}$ for $\lambda$ and $1.320$ for $n_1$ to find ${\lambda }_{n_1}$.

$\begin{array}{c}{\lambda }_{n_1}=\frac{546.1\text{nm}}{1.320}\\ =413.7\text{nm}\end{array}$

Formula to calculate the wavelength of the second beam is,

${\lambda }_{n_2}=\frac{\lambda }{n_2}$

• ${\lambda }_{n_2}$  is the wavelength of the second component
• $\lambda$ is the wavelength of the incident light
• $n_2$ is the refractive index for the  second beam

Substitute $546.1\text{nm}$ for $\lambda$ and $1.333$ for $n_2$ to find ${\lambda }_{n_3}$.

$\begin{array}{c}{\lambda }_{n_1}=\frac{546.1\text{nm}}{1.333}\\ =409.7\text{nm}\end{array}$

Thus, the first component of light has a wavelength of $413.7\text{nm}$ and the second component of light has a wavelength of $409.7\text{nm}$.

Conclusion:

The first component of light has a wavelength of $413.7\text{nm}$ and the second component of light has a wavelength of $409.7\text{nm}$.

To determine

The phase difference between the two beams when they emerge from the specimen.

Answer to Problem 61P

The phase difference between the two beams when they emerge from the specimen is $8.6°$.

Explanation of Solution

Given Info: The thickness of the specimen is $1.000\text{μm}$.

Formula to calculate the phase difference is,

$\Delta \varphi =\left(N_2-N_1\right)\left(\frac{360°}{\text{cycles}}\right)$

• $\Delta \varphi$ is the phase difference
• $N_2$ is the number of cycles of vibration of second component of light while passing through the specimen
• $N_1$ is the number of cycles of vibration of first component of light while passing through the specimen

Use $t/{\lambda }_{n_2}$ for $N_2$, $t/{\lambda }_{n_1}$ for $N_1$ in the above equation.

$\begin{array}{c}\Delta \varphi =\left(\frac{t}{{\lambda }_{n_2}}-\frac{t}{{\lambda }_{n_1}}\right)\left(\frac{360°}{\text{cycles}}\right)\\ =t\left(\frac{1}{{\lambda }_{n_2}}-\frac{1}{{\lambda }_{n_1}}\right)\left(\frac{360°}{\text{cycles}}\right)\end{array}$

• $t$ is the thickness of the specimen
• ${\lambda }_{n_2}$ is the wavelength of the second component of light while passing through the specimen
• ${\lambda }_{n_1}$ is the wavelength of the first component of light while passing through the specimen

Substitute $1.000\text{μm}$ for $t$, $413.7\text{nm}$ for ${\lambda }_{n_1}$ and $409.7\text{nm}$ for ${\lambda }_{n_2}$ to find $\Delta \varphi$.

$\begin{array}{c}\Delta \varphi =\left(\left(1.000\text{μm}\right)\left(\frac{1\text{m}}{{10}^6\text{μm}}\right)\right)\left(\frac{1}{\left(409.7\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}-\frac{1}{\left(413.7\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)\left(\frac{360°}{\text{cycles}}\right)\\ =8.6°\end{array}$

Thus, the phase difference between the two beams when they emerge from the specimen is $8.6°$.

Conclusion:

The phase difference between the two beams when they emerge from the specimen is $8.6°$.