Chapter 12, Problem 12.1P

Interpretation Introduction

(a)

Interpretation:

The frequency of the infrared light is to be calculated.

Concept introduction:

Wavelength is the distance between the two crests or troughs of the wave. It is denoted by the symbol $\left(\text{λ}\right)$. Frequency is defined as the number of waves passing through a point in unit time. It is denoted by the symbol $\left(\nu \right)$. Wavelength and frequency both are inversely proportional to each other.

Answer to Problem 12.1P

The frequency of infrared light with $\text{λ}=9\times {10}^{–6}\text{ m}$ is $3.3\times {10}^{13}{\text{ s}}^{-1}$.

Explanation of Solution

Wavelength and frequency are related to each other by the formula shown below.

$\nu =\frac{c}{\text{λ}}$…(1)

Where,

• c is the speed of light.

The value of the speed of light is $3\times {10}^8\text{m}/\text{s}$.

The wavelength of infrared light is given $\text{λ}=9\times {10}^{–6}\text{ m}$.

Substitute the value of c and $\text{λ}$ in the equation (1).

$\begin{array}{rll}\nu & =& \frac{c}{\text{λ}}\\ & =& \frac{3\times {10}^8{\text{ m s}}^{-1}}{9\times {10}^{–6}\text{ m}}\\ & =& 0.33\times {10}^{14}{\text{ s}}^{-1}\\ & \approx & 3.3\times {10}^{13}{\text{ s}}^{-1}\end{array}$

The frequency of infrared light with $\text{λ}=9\times {10}^{–6}\text{ m}$ is $3.3\times {10}^{13}{\text{ s}}^{-1}$.

Conclusion

The frequency of infrared light with $\text{λ}=9\times {10}^{–6}\text{ m}$ is calculated as $3.3\times {10}^{13}{\text{ s}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The frequency of the blue light is to be calculated.

Concept introduction:

Wavelength is the distance between the two crests or troughs of the wave. It is denoted by the symbol $\left(\text{λ}\right)$. Frequency is defined as the number of waves passing through a point in a unit time. It is denoted by the symbol $\left(\nu \right)$ Wavelength and frequency both are inversely proportional to each other.

Answer to Problem 12.1P

The frequency of blue light with $\text{λ}=4800\stackrel{\text{ο}}{\text{A}}$ is $\text{6}\text{.3}\times {\text{10}}^{14}{\text{ s}}^{-1}$.

Explanation of Solution

Wavelength and frequency are related to each other by the formula shown below.

$\nu =\frac{c}{\text{λ}}$…(1)

Where,

• c is the speed of light.

The value of the speed of light is $3\times {10}^8\text{m}/\text{s}$.

The wavelength of blue light is given $\text{λ}=4800\stackrel{\text{ο}}{\text{A}}$.

Substitute the value of c and $\text{λ}$ in the equation (1).

$\begin{array}{rll}\nu & =& \frac{c}{\text{λ}}\\ & =& \frac{3\times {10}^8{\text{ m s}}^{-1}}{4800\stackrel{\text{ο}}{\text{A}}}\end{array}$…(2)

The conversion of $1\stackrel{\text{ο}}{\text{A}}$ into metres is done as shown below.

$\begin{array}{rll}1\stackrel{\text{ο}}{\text{A}}& =& 1\times {10}^{-10}\text{ m}\\ 4800\stackrel{\text{ο}}{\text{A}}& =& 4800\times {10}^{-10}\text{ m}\end{array}$

Substitute the above equation in equation (2) as shown below.

$\begin{array}{rll}\nu & =& \frac{3\times {10}^8{\text{ m s}}^{-1}}{4800\times {10}^{-10}\text{ m}}\\ & =& \frac{3\times {10}^8{\text{ m s}}^{-1}}{4.8\times {10}^{-7}\text{ m}}\\ & =& 0.625\times {10}^{15}{\text{ s}}^{-1}\\ & \approx & \text{6}\text{.3}\times {\text{10}}^{14}{\text{ s}}^{-1}\end{array}$

The frequency of blue light with $\text{λ}=4800\stackrel{\text{ο}}{\text{A}}$ is $\text{6}\text{.3}\times {\text{10}}^{14}{\text{ s}}^{-1}$.

Conclusion

The frequency of blue light with $\text{λ}=4800\stackrel{\text{ο}}{\text{A}}$ is calculated as $\text{6}\text{.3}\times {\text{10}}^{14}{\text{ s}}^{-1}$.