Chapter 37, Problem 37.70AP

Figure CQ37.2 shows an unbroken soap film in a circular frame. The film thickness increases from lop to bottom, slowly at first and then rapidly. As a simpler model, consider a soap film (n = 1.33) contained within a rectangular wire frame. The frame is held vertically so that the film drains downward and forms a wedge with flat faces. The thickness of the film at the top is essentially zero. The film is viewed in reflected white light with near-normal incidence, and the first violet (λ = 420 nm) interference band is observed 3.00 cm from the top edge of the film, (a) Locate the first red (λ = 680 nm) interference band, (b) Determine the film thickness at the positions of the violet and red bands, (c) What is the wedge angle of the film?

To determine

The position of the first red interference band.

Answer to Problem 37.70AP

The position of the first red interference band is $4.857\text{cm}$.

Explanation of Solution

Given info: The index of refraction of the soap film is $1.33$, the thickness of the film at the top is $0$, the wavelength of the violet light is $420\text{nm}$, the position of the first violet interference band is $3.00\text{cm}$, the wavelength of the red light is $680\text{nm}$, the order of interference of red band is $0$ and the order of interference for violet band is $0$.

The diagram of thin film in rectangular wire frame is,

Figure (1)

From figure (1), the formula of tangent of angle is,

$\frac{t_{\text{r}}}{x_{\text{r}}}=\frac{t_{\text{v}}}{x_{\text{v}}}$ (1)

Here,

$t_{\text{r}}$ is the thickness of the soap film at position of red band.

$t_{\text{v}}$ is the thickness of the soap film at position of violet band.

$x_{\text{r}}$ is the position of the red interference band from the top edge of the film.

$x_{\text{v}}$ is the position of the violet interference band from the top edge of the film.

Rearrange the above equation for $x_{\text{r}}$.

$x_{\text{r}}=\left(\frac{t_{\text{r}}}{t_{\text{v}}}\right)x_{\text{v}}$ (2)

The condition for constructive interference for red band in a thin soap film is,

$2n{t}_{\text{r}}=\left(m_{\text{r}}+\frac{1}{2}\right){\lambda }_{\text{r}}$ (3)

Here,

$m_{\text{r}}$ is the order of interference of the red band.

${\lambda }_{\text{r}}$ is the wavelength of the red wave

$n$ is the refractive index of the soap film.

The condition for constructive interference for violet in a thin soap film is,

$2n{t}_{\text{v}}=\left(m_{\text{v}}+\frac{1}{2}\right){\lambda }_{\text{v}}$ (4)

Here,

$m_{\text{v}}$ is the order of interference of the violet band.

${\lambda }_{\text{v}}$ is the wavelength of the violet wave

$n$ is the refractive index of the soap film.

Divide equation (3) by (4) on both sides.

$\begin{array}{c}\frac{2n{t}_{\text{r}}}{2n{t}_{\text{v}}}=\frac{\left(m_{\text{r}}+\frac{1}{2}\right){\lambda }_{\text{r}}}{\left(m_{\text{v}}+\frac{1}{2}\right){\lambda }_{\text{v}}}\\ \frac{t_{\text{r}}}{t_{\text{v}}}=\frac{\left(m_{\text{r}}+\frac{1}{2}\right){\lambda }_{\text{r}}}{\left(m_{\text{v}}+\frac{1}{2}\right){\lambda }_{\text{v}}}\end{array}$

Substitute $0$ for $m_{\text{r}}$ and $0$ for $m_{\text{v}}$ in the above equation.

$\begin{array}{c}\frac{t_{\text{r}}}{t_{\text{v}}}=\frac{\left(0+\frac{1}{2}\right){\lambda }_{\text{r}}}{\left(0+\frac{1}{2}\right){\lambda }_{\text{v}}}\\ =\frac{{\lambda }_{\text{r}}}{{\lambda }_{\text{v}}}\end{array}$

Substitute $\frac{{\lambda }_{\text{r}}}{{\lambda }_{\text{v}}}$ for $\frac{t_{\text{r}}}{t_{\text{v}}}$ in equation (2).

$x_{\text{r}}=\left(\frac{{\lambda }_{\text{r}}}{{\lambda }_{\text{v}}}\right)x_{\text{v}}$

Substitute $680\text{nm}$ for ${\lambda }_{\text{r}}$, $420\text{nm}$ for ${\lambda }_{\text{v}}$, $3.00\text{cm}$ for $x_{\text{v}}$ in the above equation.

$\begin{array}{c}{x}_{\text{r}}=\left(\frac{680\text{nm}}{420\text{nm}}\right)\left(3.00\text{cm}\right)\\ =4.857\text{cm}\end{array}$

Conclusion:

Therefore, the position of the first red interference band is $4.857\text{cm}$.

To determine

The film thickness at the position of red and violet bands.

Answer to Problem 37.70AP

The film thickness at the position of red band is $127.82\text{nm}$ and at the position of violet bands is $78.947\text{nm}$.

Explanation of Solution

Given info: The index of refraction of the soap film is $1.33$, the thickness of the film at the top is $0$, the wavelength of the violet light is $420\text{nm}$, the position of the first violet interference band is $3.00\text{cm}$, the wavelength of the red light is $680\text{nm}$, the order of interference of red band is $0$ and the order of interference for violet band is $0$.

From equation (3), the condition for constructive interference is,

$2n{t}_{\text{r}}=\left(m_{\text{r}}+\frac{1}{2}\right){\lambda }_{\text{r}}$

Substitute $680\text{nm}$ for ${\lambda }_{\text{r}}$, $0$ for $m_{\text{r}}$, $1.33$ for $n$ in the above equation.

$\begin{array}{c}2\left(1.33\right)t_{\text{r}}=\left(0+\frac{1}{2}\right)\left(680\text{nm}\right)\\ 2.66t_{\text{r}}=\left(340\text{nm}\right)\\ t_{\text{r}}=127.82\text{nm}\end{array}$

From equation (4), the condition for constructive interference is,

$2n{t}_{\text{v}}=\left(m_{\text{v}}+\frac{1}{2}\right){\lambda }_{\text{v}}$

Substitute $420\text{nm}$ for ${\lambda }_{\text{v}}$, $0$ for $m_{\text{v}}$, $1.33$ for $n$, $420\text{nm}$ for ${\lambda }_{\text{v}}$, $3.00\text{cm}$ for $x_{\text{v}}$ in the above equation.

$\begin{array}{c}2\left(1.33\right)t_{\text{v}}=\left(0+\frac{1}{2}\right)\left(420\text{nm}\right)\\ 2.66t_{\text{v}}=\left(210\text{nm}\right)\\ t_{\text{v}}=78.947\text{nm}\end{array}$

Conclusion:

Therefore, the film thickness at the position of red band is $127.82\text{nm}$ and at the position of violet bands is $78.947\text{nm}$.

To determine

The wedge angle of the film.

Answer to Problem 37.70AP

The wedge angle of the film is $2.631\times {10}^{-6}\text{rad}$.

Explanation of Solution

Given info: The index of refraction of the soap film is $1.33$, the thickness of the film at the top is $0$, the wavelength of the violet light is $420\text{nm}$, the position of the first violet interference band is $3.00\text{cm}$, the wavelength of the red light is $680\text{nm}$, the order of interference of red band is $0$ and the order of interference for violet band is $0$.

From figure (1), the formula of tangent of angle is,

$\tan \theta =\left(\frac{t_{\text{v}}}{x_{\text{v}}}\right)$

Substitute $78.947\text{nm}$ for $t_{\text{v}}$ and $3.00\text{cm}$ for $x_{\text{v}}$ in the above equation.

$\begin{array}{c}\tan \theta =\left(\frac{78.947\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{3.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ \theta ={\tan }^{-1}\left(2.631\times {10}^{-6}\right)\\ =2.631\times {10}^{-6}\text{rad}\end{array}$

Conclusion:

Therefore, the wedge angle of the film is $2.631\times {10}^{-6}\text{rad}$.