Question

A soap film (index of refraction 1.33) is surrounded on both sides by air. When white light shines nearly perpendicularly on the film, you see bright interference colours of wavelength = 403 nm in the reflected light. What is the second smallest possible value for the thickness of the film?

Answer 1

The second smallest possible thickness of a soap film with an index of refraction of 1.33 for bright interference of 403 nm wavelength light is approximately 227 nm.

Step-by-step explanation:

Thin film interference, occurs when light waves that reflect off the top and bottom surfaces of a thin film, like a soap bubble, interfere with each other. To determine the second smallest possible thickness of the film for constructive interference of light with a wavelength of 403 nm (in the air) and index of refraction 1.33, we must consider the phase shift that occurs upon reflection. The formula to determine the thickness t for constructive interference at minimum thickness can be expressed as:

• 
  
• t = (m + 1/2) * (λ / 2*n)

where m is an integer that denotes the order of the interference, λ is the wavelength of light in the air, and n is the index of refraction of the film. For the second smallest thickness, we us m=1. Plugging in the values we get:

t = (1 + 1/2) * (403 nm / (2 * 1.33))
t ≈ 227 nm

The second smallest possible value for the thickness of the soap film to see bright interference colors of a 403 nm wavelength light is approximately 227 nm.