Question

Using the result of the problem above, calculate the distance between fringes for ${}^{\mathbf{633}\mathbf{}\mathbf{nm}}$ light falling on double slits separated by ${}^{\mathbf{0}\mathbf{.}\mathbf{0800}\mathbf{}\mathbf{mm}}$, located ${}^{\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}}$from a screen as in Figure ${}^{\mathbf{27}\mathbf{.}\mathbf{56}}$.

Short answer

The distance between the fringes is ${}^{\mathbf{2}\mathbf{.}\mathbf{37}\mathbf{}\mathbf{cm}}$.

Step-by-step solution

Given data

The wavelength used for the interference is$\mathrm{\lambda }=633\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=6.33\times {10}^{-7}\mathrm{m}$

The separation between the slits is$\mathrm{d}=0.0800\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=8.00\times {10}^{-5}\mathrm{m}$

The distance of the screen is x=3.00 m.

Evaluating the distance

Using the result of the question solved before, we then observe that: $∆\mathrm{y}=\frac{x\lambda }{\mathrm{d}}$.

We then substitute this equation using the given values in the problem.

Then, we get:

$$\begin{gathered} ∆\mathrm{y}=\frac{x\lambda }{\mathrm{d}} \\ =\frac{3.00\mathrm{m}\times \left(6.33\times {10}^{-7}\mathrm{m}\right)}{8.00\times {10}^{-5}\mathrm{m}} \\ =0.0237\mathrm{m}\left(\frac{100\mathrm{cm}}{1\mathrm{m}}\right) \\ =2.37\mathrm{cm} \end{gathered}$$

Therefore, the distance between the fringes is ${}^{2.37\mathrm{cm}}$.