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

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

- measure reactants precisely. ( first choice)

- stir to the reaction take place. ( second choice)

- wait patiently until the reaction completes. ( fourth choice)

- make sure to recover as much copper as possible. ( last choice)

would be false hope this out