A thin film with index of refraction n_2 = 1.2 lies between a piece of plastic w

Question

A thin film with index of refraction n_2 = 1.2 lies between a piece of plastic with n_3 = 1.1 and air (n_1 = 1). The thickness of the film is s = 1000 nm. White light illuminates the film from above and is reflected for some wavelengths of light in the visible spectrum (lambda = 400 nm to 700 nm). Which wavelengths correspond to the visibly reflected light? 419 nm, 618 nm 560 nm, 699 nm 436 nm, 533 nm 555 nm, 696 nm 411 nm, 650 nm The plastic with index of refraction n_3= 1.1 is changed to a material with index of refraction n_3 = 1.4. Which of the following answers best describes the wavelengths which provide visibly reflected light? 450 nm, 550 nm 420 nm, 632 nm 522 nm, 665 nm 511 nm, 699 nm 480 nm, 600 nm

Explanation / Answer

for maximum intensity

n 2s = m ( wavelength )/2

m wavelength = 4800

where m = 1,3,5,7,9,11

for range between 400nm to 700nm

m= 11 and m= 9 are possible

answer

c) 436nm , 533nm

4)

in this case

the equation becomes

n 2s = m x wavelength

m = 1,2,3,4,5,6,7

2400 = m xwavelength

here possible values are m = 5 , 4

answer

e ) 480nm , 600nm

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