Question

Coherent light with wavelength 500 nm passes through narrow slits separated by 0.340mm. At a distance from the slits large compared to their separation, what is the phase difference (in radians) in the light from the two slits at an angle of$\mathbf{23}\mathbf{.}\mathbf{0}\mathbf{°}$ from the centerline?

Short answer

The phase difference (in radians) in the light from the two slits is1669rad .

Step-by-step solution

Formula for phase difference

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{∆}\mathbf{r}\mathbf{*}\mathbf{+}\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$

Calculate the phase difference

Given:

$$\begin{gathered} \mathit{\lambda }\mathbf{=}\mathbf{500}\mathit{n}\mathit{m}\mathbf{=}\mathbf{500}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{m} \\ \mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{340}\mathit{m}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{340}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m} \end{gathered}$$

$\mathit{\theta }\mathbf{=}\mathbf{23}\mathbf{.}\mathbf{0}\mathbf{°}$

Now, the phase difference is given by

$\mathbf{\varphi }\mathbf{=}\frac{\mathbf{∆}\mathbf{r}\mathbf{*}\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$

The path difference is given by

$\mathbf{∆}\mathit{r}\mathbf{=}\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

So,

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{d}\mathbf{sin}\mathbf{\theta }\mathbf{*}\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$

Plug the given,

$\mathit{\varphi }\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{340}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{*}\mathbf{sin}\mathbf{23}\mathbf{°}\mathbf{*}\mathbf{2}\mathbf{\pi }}{\mathbf{500}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}\mathbf{=}\mathbf{1669}\mathit{r}\mathit{a}\mathit{d}$

Thus, the phase difference (in radians) in the light from the two slits is 1669rad .