Reflection by thin layers. In the figure, light is incident perpendicularly on a

Question

Reflection by thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r1 and r2 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in the table below refers to the indexes of refraction n1, n2, and n3, the type of interference, the thinlayer thickness L in nanometers, and the wavelength ? in nanometers of the light as measured in air. Where ? is missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated.

use the data for question #49, please show work and type out the final answer

Reflection by thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r1 and r2 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in the table below refers to the indexes of refraction n1, n2, and n3, the type of interference, the thinlayer thickness L in nanometers, and the wavelength ? in nanometers of the light as measured in air. Where ? is missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated. use the data for question #49, please show work and type out the final answer

Explanation / Answer

2L = [(odd #)/2] x (lambda /n2)

You have two unknowns: the odd # and lambda. However you know you need a value for the odd number which would put lambda in the range of visible light: 390 nm < visible light < 790 nm.

2L = 640 nm
640 = [(odd #)/2] x (lambda /1.75)
Now we use trial and error to find the right odd number: If odd # = 7 then lambda = 320 nm which is too low for visible light. Odd # = 3 would barely put it into the visible spectrum with 746.67 nm. I found that when I used odd # = 5 I got the right answer for my Webassign problem, so I would suggest using that.

For odd # = 5 lambda = 448 nm.

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