Given:
Thickness of soap film, d =0.0011 mm = 0.0011 × 10−3 m
Wavelength of light used,
Let the index of refraction of the soap solution be μ.
The condition of minimum reflection of light is 2μd = nλ,
where n is an interger = 1 , 2 , 3 ...
As per the question, μ has a value between 1.2 and 1.5. So,
Therefore, the index of refraction of the soap solution is 1.32.