Transmission through thin layers. In the figure, light is incident perpendicular

Question

Transmission through 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.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and r4 (the light reflects twice inside material 2). The waves of r3 and r4 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides the indexes of refraction n1, n2, and n3, the type of interference, and the wavelength in nanometers of the light as measured in air. Give the second least thickness L.

file:///Users/tjshovei/Desktop/Screen%20Shot%202016-04-22%20at%2010.27.50%20AM.png

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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