Chapter 33, Problem 27P

(a)

To determine

Thickness of the film.

Explanation of Solution

Given:

The refractive index of film is $1.50$ .

The missing wavelengths of spectrum are $360\text{nm},450\text{nm},602\text{nm}$ which means these are in destructive interface.

Formula used:

Write the expression for condition of destructive interference.

  $2t=m\frac{\lambda }{n}$

Here, $t$ is the thickness of film, $m$ is the order of spectrum, $\lambda$ is the wavelength of light and $n$ is the refractive index of film.

Rearrange the above equation for $\lambda$ .

  $\lambda =\frac{2nt}{m}$ …… (1)

Rearrange the above equation for $t$ .

  $t=\frac{m\lambda }{2n}$ …… (2)

Calculation:

Substitute $450\text{nm}$ for $\lambda$ in equation (1).

  $450\text{nm=}\frac{2nt}{m}$ …… (3)

Substitute $360\text{nm}$ for $\lambda$ and $m+1$ for $m$ due to next missing wavelength in equation (1).

  $360\text{nm=}\frac{2nt}{m+1}$ …… (4)

Divide equation (3) by equation (4).

  $\begin{array}{c}\frac{450\text{nm}}{360\text{nm}}=\frac{m+1}{m}\\ \frac{5}{4}=\frac{m+1}{m}\end{array}$

Solve for $m$ .

  $\begin{array}{c}5m=4\left(m+1\right)\\ m=4\end{array}$

Substitute $4$ for $m$ , $450\text{nm}$ for $\lambda$ and $1.50$ for $n$ in equation (2).

  $\begin{array}{c}t=\frac{\left(4\right)\left(450\text{nm}\right)}{2\left(1.50\right)}\\ =600\text{nm}\end{array}$

Conclusion:

Thus, the thickness of film is $600\text{nm}$ .

(b)

To determine

The brightest wavelength in reflected interference pattern.

Explanation of Solution

Given:

The refractive index of film is $1.50$ .

The missing wavelengths of spectrum are $360\text{nm},450\text{nm},602\text{nm}$ which means these are in destructive interface.

Formula used:

Write the expression for condition of constructive interference.

  $2t=\left(m+\frac{1}{2}\right)\frac{\lambda }{n}$

Here, $t$ is the thickness of film, $m$ is the order of spectrum, $\lambda$ is the wavelength of light and $n$ is the refractive index of film.

Rearrange the above equation for $\lambda$ .

  $\lambda =\frac{2nt}{\left(m+\frac{1}{2}\right)}$ …… (5)

Calculation:

Substitute $1.50$ for $n$ , $600\text{nm}$ for $t$ in equation (1).

  $\lambda =\frac{2\left(1.50\right)\left(600\text{nm}\right)}{\left(m+\frac{1}{2}\right)}$

  $\lambda =\frac{1800\text{nm}}{\left(m+\frac{1}{2}\right)}$ …… (6)

Substitute $0$ for $m$ in equation (6).

  $\begin{array}{c}\lambda =\frac{1800\text{nm}}{\left(0+\frac{1}{2}\right)}\\ =3600\text{nm}\end{array}$

Substitute $1$ for $m$ in equation (6).

  $\begin{array}{c}\lambda =\frac{1800\text{nm}}{\left(1+\frac{1}{2}\right)}\\ =\frac{\left(1800\text{nm}\right)}{\left(\frac{3}{2}\right)}\\ =1200\text{nm}\end{array}$

Substitute $2$ for $m$ in equation (6).

  $\begin{array}{c}\lambda =\frac{1800\text{nm}}{\left(2+\frac{1}{2}\right)}\\ =\frac{1800\text{nm}}{\left(\frac{5}{2}\right)}\\ =720\text{nm}\end{array}$

Substitute $3$ for $m$ in equation (6).

  $\begin{array}{c}\lambda =\frac{1800\text{nm}}{\left(3+\frac{1}{2}\right)}\\ =\frac{1800\text{nm}}{\left(\frac{7}{2}\right)}\\ =514\text{nm}\end{array}$

Substitute $4$ for $m$ in equation (6).

  $\begin{array}{c}\lambda =\frac{1800\text{nm}}{\left(4+\frac{1}{2}\right)}\\ =\frac{1800\text{nm}}{\left(\frac{9}{2}\right)}\\ =400\text{nm}\end{array}$

Substitute $5$ for $m$ in equation (6).

  $\begin{array}{c}\lambda =\frac{1800\text{nm}}{\left(5+\frac{1}{2}\right)}\\ =\frac{1800\text{nm}}{\left(\frac{11}{2}\right)}\\ =327\text{nm}\end{array}$

Conclusion:

Thus, the wavelengths in visible spectrum are $720\text{nm,514nm,400nm}$ .

(c)

To determine

The wavelengths in visible spectrum when refractive index of glass is $1.60$ .

Explanation of Solution

Given:

The refractive index of film is $1.60$ .

The missing wavelengths of spectrum are $360\text{nm},450\text{nm},602\text{nm}$ which means these are in destructive interface.

Formula used:

Write the expression for condition of constructive interference.

  $2t=\left(m+\frac{1}{2}\right)\frac{\lambda }{n}$

Here, $t$ is the thickness of film, $m$ is the order of spectrum, $\lambda$ is the wavelength of light and $n$ is the refractive index of film.

Rearrange the above equation for $\lambda$ .

  $\lambda =\frac{2nt}{\left(m+\frac{1}{2}\right)}$ …… (7)

Calculation:

Substitute $1.60$ for $n$ , $600\text{nm}$ for $t$ in equation (7).

  $\lambda =\frac{2\left(1.60\right)\left(600\text{nm}\right)}{\left(m+\frac{1}{2}\right)}$

  $\lambda =\frac{1920\text{nm}}{\left(m+\frac{1}{2}\right)}$ …… (8)

Substitute $0$ for $m$ in equation (8).

  $\begin{array}{c}\lambda =\frac{1920\text{nm}}{\left(0+\frac{1}{2}\right)}\\ =3840\text{nm}\end{array}$

Substitute $1$ for $m$ in equation (8).

  $\begin{array}{c}\lambda =\frac{1920\text{nm}}{\left(1+\frac{1}{2}\right)}\\ =\frac{\left(1920\text{nm}\right)}{\left(\frac{3}{2}\right)}\\ =1280\text{nm}\end{array}$

Substitute $2$ for $m$ in equation (8).

  $\begin{array}{c}\lambda =\frac{1920\text{nm}}{\left(2+\frac{1}{2}\right)}\\ =\frac{1920\text{nm}}{\left(\frac{5}{2}\right)}\\ =768\text{nm}\end{array}$

Substitute $3$ for $m$ in equation (8).

  $\begin{array}{c}\lambda =\frac{1920\text{nm}}{\left(3+\frac{1}{2}\right)}\\ =\frac{1920\text{nm}}{\left(\frac{7}{2}\right)}\\ =548.5\text{nm}\end{array}$

Substitute $4$ for $m$ in equation (8).

  $\begin{array}{c}\lambda =\frac{1920\text{nm}}{\left(4+\frac{1}{2}\right)}\\ =\frac{1920\text{nm}}{\left(\frac{9}{2}\right)}\\ =426.7\text{nm}\end{array}$

Substitute $5$ for $m$ in equation (8).

  $\begin{array}{c}\lambda =\frac{1920\text{nm}}{\left(5+\frac{1}{2}\right)}\\ =\frac{1920\text{nm}}{\left(\frac{11}{2}\right)}\\ =349\text{nm}\end{array}$

Conclusion:

Thus, the wavelengths in visible spectrum are $\text{548}\text{.5nm,426}\text{.7nm}$ .