Question

The table lists the range of wavelengths in vacuum corresponding to a given color. If light one illuminates a film that has a refractive index of 1.333 and thickness of 183 nm with white light, which color is missing from the light reflected from the film?Color Wavelength (nm)red 780 - 622orange 622 - 597yellow 597 - 577green 577 - 492blue 492 - 455violet 455 - 390
A) redB) yellowC) blueD) greenE) orange

Answer 1

Answer:

please see the answer below

Explanation:

we can use the following expression:

$\lambda=\frac{2nd}{m}$

where lambda is the wavelength of the light, n is the refractive index of the film, d is the thickens of the film and m is the order of the diffraction pattern. If m=1, we obtain the max value of lambda for the colors that are reflected.

Hence, we have:

$\lambda=\frac{2(1.333)(183nm)}{1}=487.878nm$

for m=1 487.878nm is the more reflected light. Hence, all colors with wavelength below this values are not reflected.

the following are colors that are not reflected

- 780nm red

- 622nm orange

- 597nm yellow

- 577nm greeen

- 492nm blue

hope this helps!

Answer 2

Answer:

The missing color from the light reflected is the blue.

Explanation:

The wavelength of the light is equal to:

$\lambda =2nd$

Where

n = refractive index = 1.333

d = thickness = 183 nm

Replacing:

$\lambda =2*1.33*183=486.78nm$

This value is between 455 to 492 nm (blue)