Chapter 36, Problem 23PQ

(a)

To determine

The visible wavelength that will be strongly reflected.

Answer to Problem 23PQ

The visible wavelength that will be strongly reflected is $6.50\times {10}^{-7}\text{m}$.

Explanation of Solution

Write the condition for constructive interference.

    $w=\left(m+\frac{1}{2}\right)\frac{{\lambda }_0}{2n}$

Here, $w$ is the thickness of film, ${\lambda }_{\text{0}}$ is the wavelength of the light, $n$ is the index of refraction of oil and $m$ is 0,1,2,3,…..

Rewrite the above equation to determine ${\lambda }_{\text{0}}$.

    $\begin{array}{c}w=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{\text{0}}}{2n}\\ =\frac{\left(2m+1\right){\lambda }_0}{4n}\\ {\lambda }_0=\frac{4nw}{2m+1}\end{array}$                                                                                           (I)

Conclusion:

Substitute $0$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (I) to calculate ${\lambda }_{\text{0}}$.

    $\begin{array}{c}{\lambda }_{\text{0}}=\frac{4\left(1.50\right)\left(325\text{nm}\right)}{2\left(0\right)+1}\\ =1950\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =19.5\times {10}^{-7}\text{m}\end{array}$

Substitute $1$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (I) to calculate ${\lambda }_0$.

    $\begin{array}{c}{\lambda }_{\text{0}}=\frac{4\left(1.50\right)\left(325\text{nm}\right)}{2\left(1\right)+1}\\ =650\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =6.50\times {10}^{-7}\text{m}\end{array}$

Substitute $2$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (I) to calculate ${\lambda }_{\text{0}}$.

    $\begin{array}{c}{\lambda }_{\text{0}}=\frac{4\left(1.50\right)\left(325\text{nm}\right)}{2\left(2\right)+1}\\ =390\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =3.90\times {10}^{-7}\text{m}\end{array}$

Substitute $3$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (I) to calculate ${\lambda }_{\text{0}}$.

    $\begin{array}{c}{\lambda }_{\text{o}}=\frac{4\left(1.50\right)\left(325\text{nm}\right)}{2\left(3\right)+1}\\ =279\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =2.79\times {10}^{-7}\text{m}\end{array}$

Among all four calculated wavelengths only $6.50\times {10}^{-7}\text{m}$ wavelength lies in the range of visible region.

Therefore, the visible wavelength that will be strongly reflected is $6.50\times {10}^{-7}\text{m}$.

(b)

To determine

The visible wavelength that will be strongly transmitted.

Answer to Problem 23PQ

Thevisible wavelength that will be strongly transmitted is $4.88\times {10}^{-7}\text{m}$.

Explanation of Solution

Write the condition for constructive interference under transmitted light.

    $w=\frac{m{\lambda }_{\text{0}}}{2n}$

Here, ${\lambda }_{\text{0}}$ is the wavelength of the light, $n$ is the index of refraction of oil and $m$ is 0,1,2,3,…..

Rewrite the above equation to determine ${\lambda }_{\text{0}}$.

    $\begin{array}{c}w=\frac{m{\lambda }_{\text{0}}}{2n}\\ {\lambda }_{\text{0}}=\frac{2nw}{m}\end{array}$                       (II)

Conclusion:

Substitute $1$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (II) to calculate ${\lambda }_{\text{0}}$.

    $\begin{array}{c}{\lambda }_{\text{0}}=\frac{2\left(1.50\right)\left(325\text{nm}\right)}{1}\\ =975\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =9.75\times {10}^{-7}\text{m}\end{array}$

Substitute $2$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (II) to calculate ${\lambda }_{\text{0}}$.

    $\begin{array}{c}{\lambda }_{\text{0}}=\frac{2\left(1.50\right)\left(325\text{nm}\right)}{2}\\ =488\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =4.88\times {10}^{-7}\text{m}\end{array}$

Substitute $3$ for $m$, $1.50$ for $n$ and $325\text{nm}$ for $w$ in equation (II) to calculate ${\lambda }_{\text{0}}$.

    $\begin{array}{c}{\lambda }_{\text{0}}=\frac{2\left(1.50\right)\left(325\text{nm}\right)}{3}\\ =325\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ =3.25\times {10}^{-7}\text{m}\end{array}$

Among all three calculated wavelengths, only $4.88\times {10}^{-7}\text{m}$ wavelength lies in the range of visible region.

Therefore, the visible wavelength that will be strongly transmitted is $4.88\times {10}^{-7}\text{m}$.