A plastic film with index of refraction 1.70 is put on the surface of a car wind

Question

A plastic film with index of refraction 1.70 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction 1.51. (a) What minimum thickness is required if light with wavelength 570 nm in air reflected from the two sides of the film is to interfere constructively? nm (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference? nm

Explanation / Answer

lambda = 570*10^-9 m

for constructive

2t = (m+1/2)*lambda/n

for m= 1

t = lambda/4n

t = 570*10^-9/4*1.7 = 83.82 nm

b) t = 167.64 nm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.