Thin film / interference question

[whatislifehue]

Homework Statement

Monochromatic laser light is shown through a double-slit apparatus with a slit separation of onto a screen, but no
interference pattern is seen because the thin transparent coating (index of refraction n=1.35) on the screen is creating a thin-film destructive interference effect. This is known because when the light is shown on an old uncoated screen, they saw an interference pattern with the first bright fringe deflected at an angle of 3º.Find the minimum thickness of coating that needs to be scrubbed off so that the interference pattern for that light will
appear as brightly on the screen as if it were uncoated (i.e. no destructive interference at all).

Homework Equations

$$dsin(\theta) = m\lambda \\ t = \frac{\lambda}{2n} (constructive)\\ t = \frac{\lambda}{4n} (destructive)$$

The Attempt at a Solution

We know that m = 1 at theta = 3 degrees since that's where the first interference pattern is, so the first equation becomes lambda = d*sin(theta). Plug in lambda = d*sin(theta) into t = lambda / 4*n to find the thickness at which destructive interference occurs, so t = d*sin(theta)/(4*n) (where n = 1.35) and get 106.6 nm. If 106.6nm of the coating is removed, then constructive interference on the film will occur, so 106.6nm has to be removed.

 

[ehild]

Welcome to PF!

Your derivation looks correct , but what is the distance between the slits?
And it is not sure that the coating has lambda/(4n) thickness.

 

[whatislifehue]

Welcome to PF!

Your derivation looks correct , but what is the distance between the slits?
And it is not sure that the coating has lambda/(4n) thickness.

Hi, thanks!

Woops, sorry I forgot to include the thickness. The thickness is 1.1x10^-5 m. And you make a good point that the coating might not have lambda/(4n) thickness, but the problem doesn't say how thick the layer is. It only says the minimum thickness so I assume that it is greater than 106.6nm

 

[ehild]

Your result is correct that 106.6 nm should be removed (supposing the refractive index of the screen material is greater than 1.35. Hopefully, it is.)

 

[HalfLostAndSearching]

Based on the given information, it appears that the thin transparent coating on the screen is causing destructive interference, resulting in the absence of an interference pattern. However, when the coating is removed, the interference pattern reappears. This suggests that the coating has a thickness that is causing destructive interference.

To determine the minimum thickness of the coating that needs to be removed, we can use the equation t = λ/4n, where t is the thickness of the coating, λ is the wavelength of the laser light, and n is the index of refraction of the coating. Substituting the values given in the problem, we get t = λ/(4*1.35). Since the first bright fringe is deflected at an angle of 3 degrees, we can use the equation dsin(θ) = mλ, where d is the slit separation, θ is the angle of deflection, and m is the order of the bright fringe. Since m = 1 for the first bright fringe, we can rearrange this equation to solve for λ, which gives us λ = d*sin(θ). Substituting this into our previous equation, we get t = d*sin(θ)/(4*1.35). Plugging in the values given in the problem, we get t = (0.0005 m)*sin(3 degrees)/(4*1.35) = 106.6 nm. Therefore, the minimum thickness of the coating that needs to be removed for the interference pattern to reappear as brightly as it would on an uncoated screen is 106.6 nm.

In order to achieve constructive interference, the coating needs to have a thickness equal to an integer multiple of half the wavelength of the light, which in this case is 106.6 nm. Therefore, removing this amount of coating will eliminate the destructive interference and allow the interference pattern to reappear. This demonstrates the importance of controlling the thickness of thin films in order to observe interference patterns accurately.