Chapter 24, Problem 62GP

(a)

To determine

The distance between the slits.

Answer to Problem 62GP

  $0.10mm$ .

Explanation of Solution

Given:

Wavelength of light is $\text{5}{\text{.0×10}}^{\text{-7}}\text{m}$

Distance from slits to the screen is $\text{4}\text{.0}\text{m}$ and

Distance between the fringes is $2.0\text{cm}$

Formula used:

It is known that $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Rearrange the given formula: $\sin \theta =\frac{m\lambda }{d}$ . So, $d\sin \theta =m\lambda$

It is known that $\Delta y=\frac{L\lambda \Delta m}{d}$

  $m$ is an integer

  $\lambda$ is wavelength

  $L$ is distance from slits to the screen.

  $d$ is a distance between slits

  $y$ is the distance between the fringes 

Calculation:

Path difference is multiple of wavelength (for constructive interference):

  $d\sin \theta =m\lambda ,m=0,1,2,3,\mathrm{...}$

Now, calculate the location on screen from $y=L\tan \theta$

For small angles, $\sin \theta \approx \tan \theta$ , this gives:

  $y=L\left(\frac{m\lambda }{d}\right)=\frac{mL\lambda }{d}$

For adjacent bands, $\Delta m=1$

Now, $\Delta y=\frac{L\lambda \Delta m}{d}$

Substitute the given values,

  $2.0\times {10}^{-2}m=\frac{\left(4.0m\right)\left(5.0\times {10}^{-7}m\right)\left(1\right)}{d}$

From this, $d=1.0\times {10}^{-4}m=0.10mm$

(b)

To determine

The wavelength of second source of light.

Answer to Problem 62GP

  $4.1\times {10}^{-7}m$ .

Explanation of Solution

Given:

Wavelength of light is $\text{5}{\text{.0×10}}^{\text{-7}}\text{m}$

Distance from slits to the screen is $\text{4}\text{.0}\text{m}$ and

Distance between the fringes is $2.0\text{cm}$

The distance between the slits is $\frac{1}{2}$

Formula used:

It is known that $\Delta y=\frac{L\lambda \Delta m}{d}$

  $m$ is an integer

  $\lambda$ is wavelength

  $L$ is distance from slits to the screen.

  $d$ is a distance between slits

  $y$ is the distance between the fringes 

Calculation:

Path difference is given by (for destructive interference):

  $d\sin \theta =\left(m+\frac{1}{2}\right),m=0,1,2,3,\mathrm{...}$

Again, calculate the location on screen from $y=L\tan \theta$ and use equation small angles, $\sin \theta \approx \tan \theta$ , this gives:

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

  $\frac{\left(5+\frac{1}{2}\right)L{\lambda }_2}{d}=\frac{\left(4+\frac{1}{2}\right)L{\lambda }_1}{d}$

  $\left(5+\frac{1}{2}\right){\lambda }_2=\left(4+\frac{1}{2}\right)\left(5.0\times {10}^{-7}m\right)$

Form this, ${\lambda }_2=4.1\times {10}^{-7}m$