Final answer:
The smallest thickness of a soap film for it to appear black with 540-nm light is approximately 197 nm, and the next two possible thicknesses are approximately 591 nm and 985 nm.
Step-by-step explanation:
The question pertains to the phenomenon of thin film interference, a topic in physics. To determine the smallest thickness of a soap film that appears black when illuminated with 540-nm light, we must consider that for destructive interference (which causes the black appearance) the path length difference should be an odd multiple of the light's half wavelength divided by the refractive index of the soap film (n = 1.37).
The condition for destructive interference can be written as:
- 2nt = (m + 0.5) λ₀, where m is an integer, λ₀ is the wavelength in vacuum, t is the thickness, and n is the refractive index.
For the first (smallest) thickness, we let m = 0:
- 2(1.37)t = (0 + 0.5)(540 nm)
- t = (0.5 × 540 nm) / (2 × 1.37)
- t = 197 nm (approximately)
For the next smallest possible thicknesses that show a black appearance, we let m = 1 and m = 2:
- For m = 1: t = (1.5 × 540 nm) / (2 × 1.37) = 591 nm (approximately)
- For m = 2: t = (2.5 × 540 nm) / (2 × 1.37) = 985 nm (approximately)
Therefore, the smallest thickness for the soap film to appear black is approximately 197 nm and the next two possible thicknesses are approximately 591 nm and 985 nm.