670 nm light shines through two slits that are separated by 0.3x10-3 m, and forms a striped pattern on a viewing screen. The fourth bright line on the viewing screen is located 0.05 m from the center of the pattern. What is the distance from the slits to the viewing screen in meters?

Diffraction leads to waves spreading out. If the waves spread out and end up travelling over the same space, we say they become superimposed. The superposition of two waves will result in interference.

Depending on the relative phase of the superimposed waves, we can obtain constructive or destructive interference.

If the waves add in phase they constructively interfere resulting in a larger amplitude wave (brighter region for light).
If the waves add out of phase they destructively interfere resulting in a smaller amplitude wave (darker region for light).

Thomas Young's Double-slit Experiment:

One of the most important experiments to demonstrate the wave nature of light was Thomas Young's Double-slit Experiment.

He devised an experiment in 1901 to demonstrate that light is a wave
He showed that light undergoes diffraction and interference - phenomena associated with waves.

Thomas Young allowed light to pass through a double slit barrier. This resulted in the formation of an interference pattern consisting of bright and dark fringes on an observation screen as shown in the diagram attached.

When repeating this experiment for more than two slits at a time (diffraction grating), he found that they all shared a trend, and thus, he was able to derive the condition for constructive interference:

[tex]\boxed{\huge \text{$d\sin\theta = m\lambda$}} \large \textsf{ , where:}\\\\\large \textsf{$\bullet{$ d = separation between slits$}$}\\\large \textsf{$\bullet{$ $\sf \theta$ = angle from slit to bright spot on screen$}$}\\\large \textsf{$\bullet{$ m = adjacent bright spot$}$}\\\large \textsf{$\bullet{$ $\lambda$ = wavelength of light$}$}\\[/tex]

[tex]\textsf{Note: if $\sf \theta$ is a small angle, we can assume that:}\\\\\boxed{\Large \text{$\sin \theta \approx \tan \theta = \frac{h}{L}$}} \large \textsf{ , where:\\}\\\\\large \textsf{$\bullet{$ h = separation of bright spots$}$}\\\large \textsf{$\bullet{$ L = distance between diffraction grating and detecting screen$}$}[/tex]

[tex]\Large \text{$\therefore d\frac{h}{L} = m\lambda$}\\\\\huge \text{$\Rightarrow \boxed{L = \frac{dh}{m\lambda}}$}[/tex]

Application of the formula:

[tex]\large \textsf{Applying this formula to the question:}\\\\\large \textsf{$\bullet{$ d = $\sf 0.3\times10^{-3}\ m$}$}\\\large \textsf{$\bullet{$ $\sf h = 0.05\ m$}$}\\\large \textsf{$\bullet{\sf\ m = 4}$}\\\large \textsf{$\bullet{$ $\sf \lambda=670\times10^{-9}\ m$}$}\\[/tex]

[tex]\Large \text{$\Rightarrow L = \frac{\left(0.3\times10^{-3}\right)\left(0.05\right)}{\left(4\right)\left(670\times10^{-9}\right)}$}\\\\\boxed{\boxed{\Large \text{$\therefore L = 5.60 \sf \ m$}}}[/tex]

∴ Distance from the slits to the viewing screen = 5.60 metres

To learn more about diffraction gratings:

https://brainly.com/question/32005487