Question

What wavelength monochromatic source in the visible region (390 to 710 nm) can be used to constructively reflect off a soap film (n = 1.46) if the film is 214 nm thick? lambda' = lambda/n there is a shift in wavelength in thin films. You need to use lambda' and use with thickness for path difference path difference = 2t

Answer 1

λ ≈ 624.44 nm

Step-by-step explanation:

To determine the wavelength of a monochromatic source in the visible region that can be used for constructive interference when reflecting off a soap film (with a refractive index of n = 1.46) of thickness t = 214 nm, we can use the formula for the path difference in thin films:

Path Difference = 2 * t

The path difference is crucial for constructive interference. In this case, we want to find the wavelength (λ') of the incident light in the soap film, which will lead to constructive interference.

Constructive Interference Condition:

Path Difference = m * λ' (where m is an integer)

Now, we can set up the equation:

2 * t = m * λ'

We know that λ' is related to the original wavelength (λ) and the refractive index (n) as follows:

λ' = λ / n

Rearrange the equation for λ':

λ' = λ / n

Now, we can rewrite the path difference equation using λ':

2 * t = m * λ / n

We need to solve for λ, so isolate λ:

λ = (2 * t * n) / m

Now, let's calculate λ for the given values:

t = 214 nm (thickness of the soap film)

n = 1.46 (refractive index of the soap film)

m = 1 (we want the first-order constructive interference)

λ = (2 * 214 nm * 1.46) / 1

λ ≈ 624.44 nm

So, for constructive interference with a soap film of 214 nm thickness and a refractive index of 1.46, you can use a monochromatic source with a wavelength of approximately 624.44 nm in the visible region.