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A soap bubble, when illuminated with light of frequency 5.27 Hz × 1014 Hz, appears to be especially reflective. If it is surrounded by air and if its index of refraction is 1.33, what is the thinnest thickness the soap film can be? (c = 3.00 × 108 m/s)

1 Answer

6 votes

Answer:


1.07004* 10^(-7)\ m

Step-by-step explanation:


n_s = Refractive index of bubble = 1.33

f = Frequency of light =
5.27* 10^(14)\ Hz

c = Speed of light =
3* 10^8\ m/s

The wavelength of light is given by


\lambda=(2n_st)/(m-(1)/(2))

Wavelength is also given by


\lambda=(c)/(f)

m = 1 for minimum thickness


(c)/(f)=(2n_st)/(m-(1)/(2))\\\Rightarrow t=(m-(1)/(2)c)/(2n_sf)\\\Rightarrow t=((1-(1)/(2))* 3* 10^8)/(2* 1.33* 5.27* 10^(14))\\\Rightarrow t=1.07004* 10^(-7)\ m

The minimum thickness is
1.07004* 10^(-7)\ m

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User Simao
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