A parallel beam of white light is incident normally on a water film...

For a thin film of thickness $t$ and refractive index $\mu$, the condition for constructive interference in the transmitted light (strong transmission) is given by:

$\Delta x = 2 \mu t = n \lambda$

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

Given:
$\mu = 1.33$
$t = 1.0 \times 10^{-4} \text{ cm} = 1000 \text{ nm}$

The optical path difference is:
$2 \mu t = 2 \times 1.33 \times 1000 = 2660 \text{ nm}$

Therefore, the wavelengths that are strongly transmitted are:
$\lambda = \dfrac{2660}{n} \text{ nm}$

We need $\lambda$ in the visible range ($400 \text{ nm}$ to $700 \text{ nm}$).

For $n = 4$: $\lambda = \dfrac{2660}{4} = 665 \text{ nm} \approx 666 \text{ nm}$

For $n = 5$: $\lambda = \dfrac{2660}{5} = 532 \text{ nm}$

For $n = 6$: $\lambda = \dfrac{2660}{6} = 443.3 \text{ nm} \approx 443 \text{ nm}$

Thus, the strongly transmitted wavelengths in the given options are $532 \text{ nm}$, $666 \text{ nm}$, and $443 \text{ nm}$.

Answer: $532 \text{ nm}$; $666 \text{ nm}$; $443 \text{ nm}$