Question

A 600-nm-wavelength beam of light is incident on a soap film with air on both sides. The film has an index of refraction of 1.33. Determine the film's minimum thickness to make the reflected light enhanced in brightness

Answer 1

Answer:

the film's minimum thickness to make the reflected light enhanced in brightness is 112.8 nm

Explanation:

Given the data in the question;

wavelength λ = 600 nm  

index of refraction = 1.33

so the film's minimum thickness to make the reflected light enhanced in brightness = ?

condition for maxima for reflected light is;

2nt = [ ( 2m + 1 )λ ] / 2

we solve for t

4nt = [ ( 2m + 1 )λ ]

t =  [ ( 2m + 1 )λ ] / nt

Now, for minimum thickness m = 1

t = λ / 4n

so we substitute in our values;

t = 600 nm / ( 4 × 1.33 )

t = 600 nm / 5.32

t = 112.78195 ≈ 112.8 nm  { one decimal place }

Therefore, the film's minimum thickness to make the reflected light enhanced in brightness is 112.8 nm