Chapter 24, Problem 68GP

To determine

To Describe: The diffraction pattern.

Answer to Problem 68GP

  $\sin \theta =\sin 30°-\left(\frac{m\lambda }{D}\right);m=\pm 1,\pm 2,\mathrm{...},$ .

Explanation of Solution

Given data:

The width of slit, $\text{D}$

The angles of normal, $\text{θ}=30°$

Calculation:

The path difference between top and bottom of the slit for incident wave is $\text{D}\text{sin}{\text{θ}}_{\text{i}}$ .

The path difference between top and bottom of the slit for diffracted wave is $\text{D}\text{sin}\text{θ}$ .

The net difference is zero when ${\text{θ=θ}}_i$ .

Hence, the central maximum angel at $\theta =30°$ .

When the net path difference is a multiple of a wavelength then minima will be

  $\text{D}\text{sin}{\text{θ}}_{\text{i}}-\text{D}\text{sin}\text{θ}=m\lambda ,m=\pm 1,\pm 2,\mathrm{...},$

  $\sin \theta =\sin 30°-\left\{\frac{m\lambda }{\text{D}}\right\};m=\pm 1,\pm 2,\mathrm{...},$

Conclusion: The diffracted pattern will be defined by

  $\sin \theta =\sin 30°-\left\{\frac{m\lambda }{\text{D}}\right\};m=\pm 1,\pm 2,\mathrm{...},$ .