Question

Find the slit separation of a double-slit arrangement that will produce interference fringes$\mathbf{0}\mathbf{.}\mathbf{018}\mathbf{ }\mathbf{\text{rad}}$apart on a distant screen when the light has wavelength$\mathbf{\lambda }\mathbf{=}\mathbf{589}\mathbf{ }\mathbf{\text{nm}}$.

Short answer

Thus, the wavelength of visible light is$33\mu m$.

Step-by-step solution

According to the question.

In the case of a distant screen the angle $\mathbf{\theta }$is close to zero so $\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{\approx }\mathit{\theta }$

Thus, solve further gives:

$$\begin{gathered} \Delta \theta \approx \Delta sin\theta \\ =\Delta \left(\frac{m\lambda }{d}\right) \\ =\frac{\lambda }{d}\Delta m \\ \Delta \theta =\frac{\lambda }{d} \end{gathered}$$

Here,$\mathit{\lambda }$ is wavelength of light in air/vacuum.

The wavelength of visible light. 

Use the above formula as follows:

$$\begin{gathered} d=\frac{\lambda }{\Delta \theta } \\ =\frac{589\times {10}^{-9}m}{0.018rad} \\ =3.3\times {10}^{-5}m \\ =33\mu m \end{gathered}$$

Hence, the wavelength of visible light is$33\mu m$.