Chapter 41, Problem 34P

(a)

To determine

The apparent wavelength of light.

Explanation of Solution

Given:

The actual wavelength of light is $656.3\text{nm}$ .

The distance of galaxy is $5.00\times {10}^{6}c\cdot \text{y}$ .

Formula used:

Write the expression for velocity of galaxy.

  $v=Hr$

Here, $H$ is Hubble constant, $v$ is speed of galaxy and $r$ is the distance of galaxy.

Write the expression for time dilation constant.

  $\beta =\frac{v}{c}$

Here, $c$ is speed of light, $\lambda$ is wavelength of light and $\beta$ is the time dilation constant.

Substitute $Hr$ for $v$ in above expression.

  $\beta =\frac{Hr}{c}$

Write the expression for Doppler wavelength shift for light.

  ${\lambda }^{\prime }={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$

Here, ${\lambda }^{\prime }$ is the apparent wavelength of light and ${\lambda }_0$ is the actual wavelength of light.

Substitute $\frac{Hr}{c}$ for $\beta$ in above expression.

  ${\lambda }^{\prime }={\lambda }_0\sqrt{\frac{c+Hr}{c-Hr}}$ ....... (1)

Calculation:

Substitute $656.3\text{nm}$ for ${\lambda }_0$ , $3\times {10}^5\text{km}/\text{s}$ for $c$ , $23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}$ for $H$ and $5.00\times {10}^{6}c\cdot y$ for $r$ in equation (1).

  $\begin{array}{c}{\lambda }^{\prime }=656.3\text{nm}\sqrt{\frac{\left(3\times {10}^5\text{km}/\text{s}\right)+\left(23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}\right)\left(5.00\times {10}^{6}c\cdot y\right)}{\left(3\times {10}^5\text{km}/\text{s}\right)-\left(23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}\right)\left(5.00\times {10}^{6}c\cdot y\right)}}\\ =6.6\times {10}^{-7}\text{m}\end{array}$

Conclusion:

Thus, the apparent wavelength of light is $6.6\times {10}^{-7}\text{m}$ .

(b)

To determine

The apparent wavelength of light.

Explanation of Solution

Given:

The actual wavelength of light is $656.3\text{nm}$ .

The distance of galaxy is $500\times {10}^{6}c\cdot \text{y}$ .

Formula used:

Write the expression for velocity of galaxy.

  $v=Hr$

Here, $H$ is Hubble constant, $v$ is speed of galaxy and $r$ is the distance of galaxy.

Write the expression for time dilation constant.

  $\beta =\frac{v}{c}$

Here, $c$ is speed of light, $\lambda$ is wavelength of light and $\beta$ is the time dilation constant.

Substitute $Hr$ for $v$ in above expression.

  $\beta =\frac{Hr}{c}$

Write the expression for Doppler wavelength shift for light.

  ${\lambda }^{\prime }={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$

Here, ${\lambda }^{\prime }$ is the apparent wavelength of light and ${\lambda }_0$ is the actual wavelength of light.

Substitute $\frac{Hr}{c}$ for $\beta$ in above expression.

  ${\lambda }^{\prime }={\lambda }_0\sqrt{\frac{c+Hr}{c-Hr}}$ ....... (1)

Calculation:

Substitute $656.3\text{nm}$ for ${\lambda }_0$ , $3\times {10}^5\text{km}/\text{s}$ for $c$ , $23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}$ for $H$ and $500\times {10}^{6}c\cdot y$ for $r$ in equation (1).

  $\begin{array}{c}{\lambda }^{\prime }=656.3\text{nm}\sqrt{\frac{\left(3\times {10}^5\text{km}/\text{s}\right)+\left(23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}\right)\left(500\times {10}^{6}c\cdot y\right)}{\left(3\times {10}^5\text{km}/\text{s}\right)-\left(23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}\right)\left(500\times {10}^{6}c\cdot y\right)}}\\ =6.8\times {10}^{-7}\text{m}\end{array}$

Conclusion:

Thus, the apparent wavelength of light is $6.8\times {10}^{-7}\text{m}$ .

(c)

To determine

The apparent wavelength of light.

Explanation of Solution

Given:

The actual wavelength of light is $656.3\text{nm}$ .

The distance of galaxy is $5.00\times {10}^{6}c\cdot \text{y}$ .

Formula used:

Write the expression for velocity of galaxy.

  $v=Hr$

Here, $H$ is Hubble constant, $v$ is speed of galaxy and $r$ is the distance of galaxy.

Write the expression for time dilation constant.

  $\beta =\frac{v}{c}$

Here, $c$ is speed of light, $\lambda$ is wavelength of light and $\beta$ is the time dilation constant.

Substitute $Hr$ for $v$ in above expression.

  $\beta =\frac{Hr}{c}$

Write the expression for Doppler wavelength shift for light.

  ${\lambda }^{\prime }={\lambda }_0\sqrt{\frac{1+\beta }{1-\beta }}$

Here, ${\lambda }^{\prime }$ is the apparent wavelength of light and ${\lambda }_0$ is the actual wavelength of light.

Substitute $\frac{Hr}{c}$ for $\beta$ in above expression.

  ${\lambda }^{\prime }={\lambda }_0\sqrt{\frac{c+Hr}{c-Hr}}$ ....... (1)

Calculation:

Substitute $656.3\text{nm}$ for ${\lambda }_0$ , $3\times {10}^5\text{km}/\text{s}$ for $c$ , $23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}$ for $H$ and $5.00\times {10}^{9}c\cdot y$ for $r$ in equation (1).

  $\begin{array}{c}{\lambda }^{\prime }=656.3\text{nm}\sqrt{\frac{\left(3\times {10}^5\text{km}/\text{s}\right)+\left(23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}\right)\left(5.00\times {10}^{9}c\cdot y\right)}{\left(3\times {10}^5\text{km}/\text{s}\right)-\left(23\times {10}^{-6}\text{km}\cdot {\text{s}}^{-1}\cdot {\left(c\cdot y\right)}^{-1}\right)\left(5.00\times {10}^{9}c\cdot y\right)}}\\ =9.8\times {10}^{-7}\text{m}\end{array}$

Conclusion:

Thus, the apparent wavelength of light is $9.8\times {10}^{-7}\text{m}$ .