Question

White light is sent downward onto a horizontal thin film that is sandwiched between two materials. The indexes of refraction are 1.84 for the top material, 1.72 for the thin film, and 1.55 for the bottom material. The film thickness is 5.27 × 10-7 m. Of the visible wavelengths (400 to 700 nm) that result in fully constructive interference at an observer above the film, which is the (a) longer and (b) shorter wavelength

Answer 1

Answer:

(a). The longer wavelength is 604.29 nm.

(b). The shorter wavelength is 453.2 nm.

Explanation:

Given that,

Refraction index for top material = 1.84

Refraction index of thin film = 1.72

Refraction index for bottom material = 1.55

Thickness $t=5.27\times10^{-7}\ m$

We need to calculate the wave length

Using interference in a thin film,

Constrictive interference is

$2nt=m\lambda$

$\lambda=\dfrac{2nt}{m}$

For m = 1

Put the value into the formula

$\lambda_{1}=\dfrac{2\times1.72\times5.27\times10^{-7}}{1}$

$\lambda_{1}=1812.8\ nm$

For m = 2,

Put the value into the formula

$\lambda_{1}=\dfrac{2\times1.72\times5.27\times10^{-7}}{2}$

$\lambda_{1}=906.4\ nm$

For m= 3,

Put the value into the formula

$\lambda_{1}=\dfrac{2\times1.72\times5.27\times10^{-7}}{3}$

$\lambda_{1}=604.29\ nm$

For m= 4,

Put the value into the formula

$\lambda_{1}=\dfrac{2\times1.72\times5.27\times10^{-7}}{4}$

$\lambda_{1}=453.2\ nm$

Hence, (a). The longer wavelength is 604.29 nm.

(b). The shorter wavelength is 453.2 nm.