Question

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit wide 0.105 mm. At the point in the pattern $\mathbf{3}\mathbf{.}\mathbf{25}\mathbf{°}$from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is ${\mathit{l}}_{\mathbf{0}}$?

Short answer

1. The wavelength of the radiation is 668 nm.
2. The intensity at this point is $9.36*{10}^{-5}{I}_0$.

Step-by-step solution

Formulas used to solve the question

Formula for phase difference:

$\beta =\frac{2\pi a\sin \theta }{\lambda }$

Formula for intensity:

$I=I_0(\frac{\sin \frac{\beta }{2}}{{\displaystyle \frac{\beta }{2}}}{)}^2$

Calculate the wavelength

Given: $a=0.105mm=0.105*{10}^{-3}m$

$$\begin{gathered} \theta =3.{25}^{\circ } \\ \beta =56.0rad \end{gathered}$$

Phase difference is given by

$\beta =\frac{2\pi a\sin \theta }{\lambda }$

Solve for$\lambda$,

$\lambda =\frac{2\pi a\sin \theta }{\beta }$

Plug the given,

$\lambda =\frac{2\pi *0.105*{10}^{-3*}\sin 3.{25}^{\circ }}{56.0}=6.68*{10}^{-7}m=668nm$

Calculate the intensity

The intensity is given as:

$I=I_0(\frac{\sin \frac{\beta }{2}}{{\displaystyle \frac{\beta }{2}}}{)}^2$

Plug the given,

$I=I_0(\frac{{\displaystyle \frac{\sin 56.0}{2}}}{{\displaystyle \frac{56.0}{2}}}{)}^2=9.36*{10}^{-5}{I}_0$

Thus, the wavelength of the radiation is 668 nm. The intensity at this point is $9.36*{10}^{-5}{I}_0$.