When observed using reflected light of wavelength 580 nm , a soap film with a thickness of 0.0011 mm appears dark. Determine the index of refraction of the soap solution, given that its value lies between 1.2 and 1.5 .

Step-by-step solution

For a soap film of thickness t and refractive index in air, light reflected from the top surface undergoes a phase change of , while light reflected from the bottom surface undergoes no phase change. The condition for destructive interference (dark film) in reflected light at normal incidence is given by: 2 t = n where n = 1, 2, 3, Rearranging the formula to solve for the refractive index , we get: = n 2 t Substituting the given values, = 580 nm = 580 10⁻⁹ m and t = 0.0011 mm = 1100 10⁻⁹ m : = n 580 10⁻⁹ 2 1100 10⁻