Chapter 36, Problem 22P

(a)

To determine

The wavelength and the color of the light in the visible spectrum most strongly reflected.

Answer to Problem 22P

The value of wavelength of the light for $m=1$ is $541.33\text{nm}$ and the colour of the light in the visible spectrum most strongly reflected is green.

Explanation of Solution

Given Information: The refractive index of the oil film is $1.45$, thickness of the film is $280\text{nm}$.

It is given that an oil film floating on water is illuminated by white light at normal incidence as shown in figure given below.

Figure (1)

For most strongly reflected waves:

Write the expression for the constructive interference in thin film.

$2{\mu }_{\text{oil}}t=\left(m+\frac{1}{2}\right)\lambda$ (1)

Here,

${\mu }_{\text{oil}}$ is the refractive index of the oil film.

$\lambda$ is the value of wavelength of the light.

$t$ is the thickness of the film.

$m$ is the order number.

From equation (1), formula to calculate the value of wavelength of the light is,

$\lambda =\frac{2{\mu }_{\text{oil}}t}{\left(m+\frac{1}{2}\right)}$ (2)

From equation (2), formula to calculate the value of wavelength of the light for $m=0$ is,

${\lambda }_0=\frac{2{\mu }_{\text{oil}}t}{\left(m+\frac{1}{2}\right)}$ (3)

Here,

${\lambda }_0$ is the value of wavelength of the light for $m=0$.

Substitute $0$ for $m$, $1.45$ for ${\mu }_{\text{oil}}$, $280\text{nm}$ for $t$ in equation (3) to find ${\lambda }_0$,

$\begin{array}{c}{\lambda }_0=\frac{2\times 1.45\times 280\text{nm}}{\left(0+\frac{1}{2}\right)}\\ =1624\text{nm}\end{array}$

The range for the wavelength of the visible light is $390\text{nm}$ to $700\text{nm}$.

Thus, the value of wavelength of the light for $m=0$ is $1624\text{nm}$ and the colour of the light in the invisible spectrum is infared.

From equation (2), formula to calculate the value of wavelength of the light for $m=1$ is,

${\lambda }_1=\frac{2{\mu }_{\text{oil}}t}{\left(m+\frac{1}{2}\right)}$ (4)

Here,

${\lambda }_1$ is the value of wavelength of the light for $m=1$.

Substitute $1$ for $m$, $1.45$ for ${\mu }_{\text{oil}}$, $280\text{nm}$ for $t$ in equation (4) to find ${\lambda }_1$,

$\begin{array}{c}{\lambda }_1=\frac{2\times 1.45\times 280\text{nm}}{\left(1+\frac{1}{2}\right)}\\ =541.33\text{nm}\end{array}$

Thus, the value of wavelength of the light for $m=1$ is $541.33\text{nm}$ and the colour of the light in the visible spectrum is green.

From equation (2), formula to calculate the value of wavelength of the light for $m=2$ is,

${\lambda }_2=\frac{2{\mu }_{\text{oil}}t}{\left(m+\frac{1}{2}\right)}$ (4)

Here,

${\lambda }_2$ is the value of wavelength of the light for $m=2$.

Substitute $1$ for $m$, $1.45$ for ${\mu }_{\text{oil}}$, $280\text{nm}$ for $t$ in equation (4) to find ${\lambda }_2$,

$\begin{array}{c}{\lambda }_2=\frac{2\times 1.45\times 280\text{nm}}{\left(2+\frac{1}{2}\right)}\\ =324.8\text{nm}\\ \approx 325\text{nm}\end{array}$

Thus, the value of wavelength of the light for $m=2$ is $325\text{nm}$ and the colour of the light in the invisible spectrum is ultraviolet.

Conclusion:

Therefore, the value of wavelength of the light for $m=1$ is $541.33\text{nm}$ and the colour of the light in the visible spectrum most strongly reflected is green.

(b)

To determine

The wavelength and the color of the light in the spectrum most strongly transmitted.

Answer to Problem 22P

The value of wavelength of the light for $m=1$ is $271\text{nm}$ and the colour of the light in the visible spectrum most strongly transmitted is violet.

Explanation of Solution

Given Information: The refractive index of the oil film is $1.45$, thickness of the film is $280\text{nm}$.

For most strongly transmitted waves:

Write the expression for the destructive interference in thin film.

$2{\mu }_{\text{oil}}t=m\lambda$ (5)

From equation (5), formula to calculate the value of wavelength of the light is,

$\lambda =\frac{2{\mu }_{\text{oil}}t}{m}$ (6)

From equation (6), formula to calculate the value of wavelength of the light for $m=1$ is,

${\lambda }_0=\frac{2{\mu }_{\text{oil}}t}{m}$ (7)

Here,

${\lambda }_0$ is the value of wavelength of the light for $m=1$.

Substitute $1$ for $m$, $1.45$ for ${\mu }_{\text{oil}}$, $280\text{nm}$ for $t$ in equation (7) to find ${\lambda }_0$,

$\begin{array}{c}{\lambda }_0=\frac{2\times 1.45\times 280\text{nm}}{1}\\ =812\text{nm}\end{array}$

Thus, the value of wavelength of the light for $m=1$ is $812\text{nm}$ and the colour of the light in the invisible spectrum is infared.

From equation (6), formula to calculate the value of wavelength of the light for $m=2$ is,

${\lambda }_1=\frac{2{\mu }_{\text{oil}}t}{m}$ (8)

Here,

${\lambda }_1$ is the value of wavelength of the light for $m=2$.

Substitute $2$ for $m$, $1.45$ for ${\mu }_{\text{oil}}$, $280\text{nm}$ for $t$ in equation (8) to find ${\lambda }_1$,

$\begin{array}{c}{\lambda }_1=\frac{2\times 1.45\times 280\text{nm}}{2}\\ =406\text{nm}\end{array}$

Thus, the value of wavelength of the light for $m=2$ is $406\text{nm}$ and the colour of the light in the visible spectrum is violet.

From equation (6), formula to calculate the value of wavelength of the light for $m=3$ is,

${\lambda }_2=\frac{2{\mu }_{\text{oil}}t}{m}$ (9)

Here,

${\lambda }_2$ is the value of wavelength of the light for $m=3$.

Substitute $3$ for $m$, $1.45$ for ${\mu }_{\text{oil}}$, $280\text{nm}$ for $t$ in equation (9) to find ${\lambda }_2$,

$\begin{array}{c}{\lambda }_2=\frac{2\times 1.45\times 280\text{nm}}{3}\\ =270.666\text{nm}\\ \approx 271\text{nm}\end{array}$

Thus, the value of wavelength of the light for $m=3$ is $271\text{nm}$ and the colour of the light in the invisible spectrum is ultraviolet.

Conclusion:

Therefore, the value of wavelength of the light for $m=1$ is $271\text{nm}$ and the colour of the light in the visible spectrum most strongly transmitted is violet.