Question

You want to coat a pane of glass that has an index of refraction of 1.54 with a 195-nm-thick layer of material that has an index of refraction greater than that of the glass. The purpose of the coating is to cancel light reflected off the top of the film having a wavelength (in air) of 540 nm. The coating material is very expensive, so you want to use the thinnest possible layer. 1) What should be the index of refraction of the coating? (Express your answer to two significant figures.) Submit Your submissions: 0.69 X Computed value: 0.69 Submitted: Wednesday, February 5 at 7:21 PM Feedback: 2) If, due to technological difficulties, you cannot achieve a uniform coating at the desired thickness, what is the next thickness of the coating you could use? (Express your answer to three significant figures.) nm Submit 3) If, due to technological difficulties, you cannot achieve a uniform coating at the desired thickness, what is the second next thickness of the coating you could use? (Express your answer to three significant figures.) nm Submit 4) If, due to technological difficulties, you cannot achieve a uniform coating at the desired thickness, what is the third next thickness of the coating you could use? (Express your answer to three significant figures.) nm Submit

Answer 1

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195 * here d= Bhickness = tesom 27 WED = 540 nm mampilo 1.54 195 - 2x mailom 95x108 1 x 5411 o ... alilor = 2.13 under of refraction of coating 1-54 28 THU 54 009 Now and F. m m 2 or 2x 8-13 x d = x 540x109 do 390.42X169m Z 390.42 nm next buickness. Next luckness Qx_2-13 x da 3x540x10 on, do 585.63 mm 2x2.13 d = 4 x 540 x109. g da 780.65 nm Noter :59