Question

You dip a wire loop into soapy water (n = 1.33) and hold it up vertically to look at the soap film in white light. The soap film looks dark at the top because it has sagged, and its thickness there is nearly zero, causing the reflected wavelengths to interfere destructively. Partway down the loop you see the first red band of the reflected white light. What is the thickness of the soap film there? (Take the wavelength of red light to be 680 nm.)

Answer 1

Answer:

the thickness of the soap film is 127.82 nm or 130 nm

Explanation:

Given the data in the question;

n = 1.33

λ$_t$ = 680 nm = 680 × 10⁻⁹ m

m = 1 and β = 0

When we see the red fringe, its a point of maximum reflection

hence, for interference with a thin soap film, we say;

2 × n × d × cos( β ) = ( m - 0.5) × λ$_t$

so we substitute in our given values;

2 × 1.33 × d × cos( 0 ) = ( 1 - 0.5) × ( 680 × 10⁻⁹ )

2.66 × cos( 0 ) × d = 0.5 × ( 680 × 10⁻⁹ )

2.66 × 1 × d = 3.4 × 10⁻⁷

d = ( 3.4 × 10⁻⁷ ) / 2.66

d = 127.82 × 10⁻⁹ m

d = 127.82 nm ≈ 130 nm

Therefore, the thickness of the soap film is 127.82 nm or 130 nm