Question

Light is incident on a grating at an angle c as shown in Fig. 36-49.

Show that bright fringes occur at angles $\mathit{\theta }$ that satisfy the equation

$\mathit{d}\mathbf{(}\mathbf{sin\psi }\mathbf{+}\mathbf{sin\theta }\mathbf{)}\mathbf{=}\mathit{m}\mathit{\lambda }$m=0, 1, 2, 3(Compare this equation with Eq. 36-25.) Only the special case$\mathit{\psi }\mathbf{=}\mathbf{0}$has been treated in this chapter

Short answer

It is proved that the equation$d\left(\sin \psi +\sin \theta \right)=m\lambda$ have bright fringes occur at angle $\theta$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The equation for bright fringe is,$d\left(\sin \psi +\sin \theta \right)=m\lambda$

Concept/Significance of diffraction grating.

A diffraction grating divides light into wavelength-based components. It has minute, typically periodic features that cause the incident light's angle to be warped.

Bright fringes occur at angles

Drawing perpendiculars from the top ray's direction change (point P) to the incident and diffracted paths of the lower one will reveal the additional distance travelled by the lower one. Points A and C on the bottom ray's trajectories denote where these perpendiculars cross. Point B is the location of the bottom ray's direction shift. Keep in mind that angle BPC and angle APB are equivalent to B$\psi$ and $\theta$, respectively.

The path difference two adjacent rays is given by,

$$\begin{gathered} \Delta x=\left|AB\right|+\left|BC\right| \\ =d\sin \psi +d\sin \theta \end{gathered}$$

The condition of bright fringe is given by,

$$\begin{gathered} \Delta x=d\left(\sin \psi +\sin \theta \right) \\ =m\lambda \end{gathered}$$

Here, $m=0,1,2.....$

When $\psi =0$approaches above equation turns into brags law.

Thus, it is proved that the equation $d\left(\sin \psi +\sin \theta \right)=m\lambda$have bright fringes occur at angles $\theta$.