Chapter 33, Problem 37P

(a)

To determine

The spacing between successive maxima near the center fringe.

Explanation of Solution

Given:

The wavelength of the light used is $500\text{nm}$ .

The separation between the slits is $1\text{cm}$ .

The distance between slits and the screen is $1\text{m}$ .

Formula used:

Write the expression for the distance on the screen to the $m^{\text{th}}$ bright fringe.

  $y_{\text{m}}=m\frac{\lambda L}{d}$   ...... (1)

Here $y_{\text{m}}$ is the distance for $m^{\text{th}}$ bright fringe, $\lambda$ is the wavelength of the light used, $L$ is the distance between slits and the screen, $d$ is theseparation between the slits and $m$ is an integer.

Write the expression for the distance on the screen to the ${\left(m+1\right)}^{\text{th}}$ bright fringe.

  $y_{\text{m+1}}=\left(m+1\right)\frac{\lambda L}{d}$   ...... (2)

Here $y_{\text{m+1}}$ is the distance for ${\left(m+1\right)}^{\text{th}}$ bright fringe

Subtract equation (2) from equation (1).

  $\begin{array}{c}\Delta y=y_{\text{m+1}}-y_{\text{m}}\\ =\left(m+1\right)\frac{\lambda L}{d}-m\frac{\lambda L}{d}\\ =\frac{\lambda L}{d}\end{array}$   ...... (3)

Here $\Delta y$ is the separation between ${\left(m+1\right)}^{\text{th}}$ and $m^{\text{th}}$ bright fringe.

Calculation:

Substitute $500\text{nm}$ for $\lambda$ , $1\text{cm}$ for $d$ and $1\text{m}$ for $L$ in equation (3).

  $\begin{array}{c}\Delta y=\left(\frac{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(1\text{m}\right)}{1\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ =50\left({10}^{-6}\right)\text{m}\\ =50\left({10}^{-6}\right)\left(\frac{\text{μm}}{{10}^{-6}\text{m}}\right)\text{m}\\ =50\text{μm}\end{array}$

Conclusion:

Thus, the spacing between successive maxima near the center fringe is $50\text{μm}$ .

(b)

To determine

The observation of the interference of the light on the screen for the given arrangement.

Explanation of Solution

Given:

The wavelength of the light used is $500\text{nm}$ .

The separation between the slits is $1\text{cm}$ .

The distance between slits and the screen is $1\text{m}$ .

Formula used:

Write the expression for the distance on the screen to the $m^{\text{th}}$ bright fringe.

  $y_{\text{m}}=m\frac{\lambda L}{d}$   ...... (1)

Here $y_{\text{m}}$ is the distance for $m^{\text{th}}$ bright fringe, $\lambda$ is the wavelength of the light used, $L$ is the distance between slits and the screen, $d$ is theseparation between the slits and $m$ is an integer.

Write the expression for the distance on the screen to the ${\left(m+1\right)}^{\text{th}}$ bright fringe.

  $y_{\text{m+1}}=\left(m+1\right)\frac{\lambda L}{d}$   ...... (2)

Here $y_{\text{m+1}}$ is the distance for ${\left(m+1\right)}^{\text{th}}$ bright fringe

Subtract equation (2) from equation (1).

  $\begin{array}{c}\Delta y=y_{\text{m+1}}-y_{\text{m}}\\ =\left(m+1\right)\frac{\lambda L}{d}-m\frac{\lambda L}{d}\\ =\frac{\lambda L}{d}\end{array}$   ...... (3)

Here $\Delta y$ is the separation between ${\left(m+1\right)}^{\text{th}}$ and $m^{\text{th}}$ bright fringe.

Write the expression for Rayleigh criterion.

  $\theta =1.22\frac{\lambda }{D}$   ...... (4)

Here $\theta$ is the angle of resolution and $D$ is the diameter of the slit opening.

Calculation:

Substitute $500\text{nm}$ for $\lambda$ , $1\text{cm}$ for $d$ and $1\text{m}$ for $L$ in equation (3).

  $\begin{array}{c}\Delta y=\left(\frac{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(1\text{m}\right)}{1\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\right)\\ =50\left({10}^{-6}\right)\text{m}\\ =50\left({10}^{-6}\right)\left(\frac{\text{μm}}{{10}^{-6}\text{m}}\right)\text{m}\\ =50\text{μm}\end{array}$

For simplicity, take $D=\Delta y$

Substitute $50\text{μm}$ for $D$ and $500\text{nm}$ for $\lambda$ in equation (4).

  $\begin{array}{c}\theta =1.22\left(\frac{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{50\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\right)\\ =\frac{1.22}{100}\text{rad}\\ =0.0122\text{rad}\end{array}$

Conclusion:

Thus, the interference of the light on the screen for the given arrangement will be observed of about $0.0122\text{rad}$ , which is very less.

(c)

To determine

The distance between the slits placed for the maxima.

Explanation of Solution

Given:

The wavelength of the light used is $500\text{nm}$ .

The distance between slits and the screen $1\text{m}$ .

The separation between ${\left(m+1\right)}^{\text{th}}$ and $m^{\text{th}}$ bright fringe is $1\text{mm}$ .

Formula used:

Write the expression for the distance on the screen to the $m^{\text{th}}$ bright fringe.

  $y_{\text{m}}=m\frac{\lambda L}{d}$   ...... (1)

Here $y_{\text{m}}$ is the distance for $m^{\text{th}}$ bright fringe, $\lambda$ is the wavelength of the light used, $L$ is the distance between slits and the screen, $d$ is theseparation between the slits and $m$ is an integer.

Write the expression for the distance on the screen to the ${\left(m+1\right)}^{\text{th}}$ bright fringe.

  $y_{\text{m+1}}=\left(m+1\right)\frac{\lambda L}{d}$   ...... (2)

Here $y_{\text{m+1}}$ is the distance for ${\left(m+1\right)}^{\text{th}}$ bright fringe

Subtract equation (2) from equation (1).

  $\begin{array}{c}\Delta y=y_{\text{m+1}}-y_{\text{m}}\\ =\left(m+1\right)\frac{\lambda L}{d}-m\frac{\lambda L}{d}\\ =\frac{\lambda L}{d}\end{array}$   ...... (3)

Here $\Delta y$ is the separation between ${\left(m+1\right)}^{\text{th}}$ and $m^{\text{th}}$ bright fringe.

Simplify the above expression for separation between the slits.

  $d=\frac{\lambda L}{\Delta y}$   ...... (4)

Calculation:

Substitute $500\text{nm}$ for $\lambda$ , $1\text{mm}$ for $\Delta y$ and $1.00\text{m}$ for $L$ in equation (4).

  $\begin{array}{c}d=\left(\frac{500\text{nm}\left(\frac{{10}^{-6}\text{mm}}{1\text{nm}}\right)1.00\text{m}\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)}{1\text{mm}}\right)\\ =0.500\text{mm}\end{array}$

Conclusion:

Thus, the distance between the slits is $0.500\text{mm}$ .