A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm.. What is the index of refraction of the soap solution, if it is known to be between 1-2 and 1.5 ?

Explanation

To solve the problem, we need to determine the index of refraction (μ) of the soap film given its thickness and the wavelength of light. The soap film appears dark when viewed in reflected light, which indicates that destructive interference is occurring.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Thickness of the soap film (t) = 0.0011 mm = 0.0011 × 10^-3 m = 1.1 × 10^-6 m
- Wavelength of light (λ) = 580 nm = 580 × 10^-9 m = 5.8 × 10^-7 m

2. **Understanding Path Difference:**
- For a soap film, when light reflects off the top and bottom surfaces, the path difference (Δx) for destructive interference is given by:
\[ \Delta x = 2μt - \frac{λ}{2} \]
- Here, the term \(\frac{λ}{2}\) accounts for the phase change of π (or half a wavelength) that occurs upon reflection from a medium of higher refractive index.

3. **Condition for Dark Fringe:**
- For dark fringes (destructive interference), the path difference must equal:
\[ \Delta x = (2n - 1) \frac{λ}{2} \]
- Where n is an integer (order of the dark fringe).

4. **Setting the Equations Equal:**
- We equate the two expressions for path difference:
\[ 2μt - \frac{λ}{2} = (2n - 1) \frac{λ}{2} \]
- Rearranging gives:
\[ 2μt = (2n - 1) \frac{λ}{2} + \frac{λ}{2} \]
\[ 2μt = nλ \]

5. **Solving for μ:**
- From the above equation, we can express μ as:
\[ μ = \frac{nλ}{2t} \]

6. **Substituting Values:**
- Substitute the values of λ and t into the equation:
\[ μ = \frac{n \times 580 \times 10^{-9}}{2 \times 1.1 \times 10^{-6}} \]
\[ μ = \frac{n \times 580 \times 10^{-9}}{2.2 \times 10^{-6}} \]
\[ μ = \frac{n \times 580}{2.2} \times 10^{-3} \]
\[ μ = 0.263636n \]

7. **Finding n:**
- We know that μ must lie between 1.2 and 1.5:
\[ 1.2 < 0.263636n < 1.5 \]
- Dividing the entire inequality by 0.263636:
\[ \frac{1.2}{0.263636} < n < \frac{1.5}{0.263636} \]
\[ 4.55 < n < 5.68 \]
- Since n must be an integer, the only possible value for n is 5.

8. **Calculating μ:**
- Substitute n = 5 back into the equation for μ:
\[ μ = 0.263636 \times 5 = 1.31818 \]

### Conclusion:
The index of refraction of the soap solution is approximately **1.32**, which lies between 1.2 and 1.5.