Question

A $\mathbf{600}\mathbf{}\mathbf{nm}$-thick soap film $\left(\mathbf{n}=\mathbf{1}.\mathbf{40}\right)$in air is illuminated with white light in a direction perpendicular to the film. For how many different wavelengths in the $\mathbf{300}$to $\mathbf{700}\mathbf{}\mathbf{nm}$ range is there (a) fully constructive interference and (b) fully destructive interference in the reflected light?

Short answer

(a) The number of wavelengths lying in visible range for constructive interference by thick soap film is 4 .

(b) The number of wavelengths lying in visible range for destructive interference by thick soap film is 5 .

Step-by-step solution

Identification of given data

The index of refraction for thick soap film is $n=1.40$

The range of wavelength is ${\lambda }_{\min }-{\lambda }_{\max }=300\text{nm}-700\text{nm}$.

The thickness of soap film is $t=600\text{nm}$.

Understanding the concept

The index of refraction is the ratio of the speed of light in medium to speed of light in vacuum.

(a) Determination of order of fringe for minimum and maximum wavelength of visible range for constructive interference

The minimum order of fringe for destructive interference of thick soap film in the visible range is given as:

$m_{\min }+\frac{1}{2}=\frac{2nt}{{\lambda }_{\min }}$

Substitute all the values in the equation.

$$\begin{gathered} m_{\min }+\frac{1}{2}=\frac{2\left(1.40\right)\left(600\text{nm}\right)}{300\text{nm}} \\ {m}_{\min }=5.6-0.5 \\ {m}_{\min }=5 \end{gathered}$$

The maximum order of fringe for destructive interference of thick soap film in visible range is given as:

$m_{\max }+\frac{1}{2}=\frac{2nt}{{\lambda }_{\max }}$

Substitute all the values in equation.

$$\begin{gathered} m_{\max }+\frac{1}{2}=\frac{2\left(1.40\right)\left(600\text{nm}\right)}{700\text{nm}} \\ {m}_{\max }=2.4-0.5 \\ {m}_{\max }=2 \end{gathered}$$

The second, third, fourth and fifth wavelengths lie in visible range for constructive interference by thick soap film.

Therefore, the number of wavelengths lying in visible range for constructive interference by thick soap film is 4 .

(b) Determination of order of fringe for minimum and maximum wavelength of visible range for destructive interference

The minimum order of fringe for destructive interference of thick soap film in the visible range is given as:

$m_{\min }=\frac{2nt}{{\lambda }_{\min }}$

Substitute all the values in the equation.

$\begin{array}{rcl}{m}_{\min }& =& \frac{2\left(1.40\right)\left(600\text{nm}\right)}{300\text{nm}}\\ & =& 5.6\\ & \approx & 6\end{array}$

The maximum order of fringe for destructive interference of thick soap film in visible range is given as:

$m_{\max }=\frac{2nt}{{\lambda }_{\max }}$

Substitute all the values in equation.

$\begin{array}{rcl}d{m}_{\max }& =& \frac{2\left(1.40\right)\left(600\text{nm}\right)}{700\text{nm}}\\ & =& 2.4\\ & \approx & 2\end{array}$

The second, third, fourth, fifth and sixth wavelengths lie in visible range for destructive interference by thick soap film.

Therefore, the number of wavelengths lying in visible range for destructive interference by thick soap film is 5.