When observed using reflected light of wavelength 580 nm , a soap...

For a soap film of thickness $t$ and refractive index $\mu$ in air, light reflected from the top surface undergoes a phase change of $\pi$, 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:

$2 \mu t = n \lambda$

where $n = 1, 2, 3, \dots$

Rearranging the formula to solve for the refractive index $\mu$, we get:

$\mu = \dfrac{n \lambda}{2 t}$

Substituting the given values, $\lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m}$ and $t = 0.0011 \text{ mm} = 1100 \times 10^{-9} \text{ m}$:

$\mu = \dfrac{n \times 580 \times 10^{-9}}{2 \times 1100 \times 10^{-9}} = \dfrac{580 n}{2200} = \dfrac{29 n}{110}$

Evaluating $\mu$ for different integer values of $n$:

For $n = 4$, $\mu = \dfrac{29 \times 4}{110} \approx 1.05$

For $n = 5$, $\mu = \dfrac{29 \times 5}{110} = \dfrac{145}{110} \approx 1.32$

For $n = 6$, $\mu = \dfrac{29 \times 6}{110} = \dfrac{174}{110} \approx 1.58$

Since the refractive index lies between $1.2$ and $1.5$, the valid value is obtained for $n = 5$, which gives $\mu \approx 1.32$.