Question

Light of wavelength 5800A°falling at nearly normal incidence gets reflected from a soap film of refractive index 1.4.what is the least thickness of the film that will appear bright and dark?

Answer 1

Answer on Question #81221 Physics / Optics

**Question.** Light of wavelength 5800 Å falling at nearly normal incidence gets reflected from a soap film of refractive index 1.4. what is the least thickness of the film that will appear bright and dark?

Solution.

For bright

$$2dn \cos r \pm \frac{\lambda_0}{2} = 2d\sqrt{n^2 - \sin^2 i} \pm \frac{\lambda_0}{2} = m\lambda_0 (m = 1, 2, \dots).$$

For dark

$$2dn \cos r \pm \frac{\lambda_0}{2} = 2d\sqrt{n^2 - \sin^2 i} \pm \frac{\lambda_0}{2} = (2m + 1)\frac{\lambda_0}{2} (m = 1, 2, \dots).$$

In our case $n > n_0$, where $n_0 = 1$ (air). So,

For bright

$$\begin{aligned} 2dn \cos r - \frac{\lambda_0}{2} &= \lambda_0 \rightarrow 2dn \cos 0{}^\circ - \frac{\lambda_0}{2} = \lambda_0 \rightarrow 2dn = \frac{3}{2}\lambda_0 \rightarrow d = \frac{3}{4} \cdot \frac{\lambda_0}{n} \\ &= \frac{3}{4} \cdot \frac{5800 \cdot 10^{-10}}{1.4} = 3.107 \cdot 10^{-7} \, m. \end{aligned}$$

For dark

$$\begin{aligned} 2dn \cos r - \frac{\lambda_0}{2} &= \frac{\lambda_0}{2} \rightarrow 2dn \cos 0{}^\circ - \frac{\lambda_0}{2} = \frac{\lambda_0}{2} \rightarrow 2dn = \lambda_0 \rightarrow d = \frac{\lambda_0}{2n} = \frac{5800 \cdot 10^{-10}}{2 \cdot 1.4} \\ &= 2.07 \cdot 10^{-7} \, m. \end{aligned}$$

**Answer.** For bright $-d = 3.107 \cdot 10^{-7} \, m$; for dark $-d = 2.07 \cdot 10^{-7} \, m$.

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