Question

both sides of the film is air. The angle of incidence is 60°. Find the minimum film thickness if reflected light is most intense for $\lambda$ = 600 nm.

• A. The condition for constructive interference (most intense reflected light) in a thin film, considering a phase change of $\pi$ at the first interface (air-film, as $\mu_{film} > \mu_{air}$) and no phase change at the second interface (film-air), is given by: $2\mu t \cos r = \frac{(2m-1)\lambda}{2}$ where $\mu$ is the refractive index of the film, $t$ is the film thickness, $r$ is the angle of refraction, $\lambda$ is the wavelength of light, and $m$ is an integer. For the minimum film thickness, we take $m=1$: $2\mu t \cos r = \frac{\lambda}{2}$.
• B. Using Snell's Law ($n_1 \sin i = n_2 \sin r$): $1 \cdot \sin 60^\circ = 1.5 \sin r$. $\frac{\sqrt{3}}{2} = \frac{3}{2} \sin r$. $\sin r = \frac{\sqrt{3}}{3}$.
• C. Now, calculate $\cos r$: $\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \left(\frac{\sqrt{3}}{3}\right)^2} = \sqrt{1 - \frac{3}{9}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.
• D. Substitute the values into the condition for minimum thickness: $2 \cdot (1.5) \cdot t \cdot \left(\frac{\sqrt{2}}{\sqrt{3}}\right) = \frac{600 \text{ nm}}{2}$. $3 \cdot t \cdot \frac{\sqrt{2}}{\sqrt{3}} = 300 \text{ nm}$. $t \cdot \frac{3\sqrt{2}}{\sqrt{3}} = 300 \text{ nm}$. $t \cdot \sqrt{3}\sqrt{2} = 300 \text{ nm}$. $t \sqrt{6} = 300 \text{ nm}$. $t = \frac{300}{\sqrt{6}} \text{ nm}$. Rationalizing the denominator: $t = \frac{300\sqrt{6}}{6} \text{ nm} = 50\sqrt{6} \text{ nm}$.

Answer 1

Answer: (D)

The minimum film thickness is $50\sqrt{6}$ nm.

Answer 2

The condition for constructive interference in a thin film when light is reflected from a medium of lower refractive index to a medium of higher refractive index (air to film) is $2\mu t \cos r = (m - \frac{1}{2})\lambda$. Since the light is reflected from the film to air, there is no phase change at the second interface. For maximum intensity (constructive interference), the condition is $2\mu t \cos r = (2m-1)\frac{\lambda}{2}$. For minimum thickness, $m=1$, so $2\mu t \cos r = \frac{\lambda}{2}$.

Given: Refractive index of the film, $\mu = 1.5$ Angle of incidence, $i = 60^\circ$ Wavelength, $\lambda = 600$ nm

Using Snell's Law: $n_1 \sin i = n_2 \sin r$ $1 \cdot \sin 60^\circ = 1.5 \sin r$ $\frac{\sqrt{3}}{2} = \frac{3}{2} \sin r$ $\sin r = \frac{\sqrt{3}}{3}$

Now, find $\cos r$: $\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \left(\frac{\sqrt{3}}{3}\right)^2} = \sqrt{1 - \frac{3}{9}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$

Substitute the values into the condition for minimum thickness: $2 \cdot (1.5) \cdot t \cdot \left(\frac{\sqrt{2}}{\sqrt{3}}\right) = \frac{600 \text{ nm}}{2}$ $3 \cdot t \cdot \frac{\sqrt{2}}{\sqrt{3}} = 300 \text{ nm}$ $t \cdot \frac{3\sqrt{2}}{\sqrt{3}} = 300 \text{ nm}$ $t \cdot \sqrt{3}\sqrt{2} = 300 \text{ nm}$ $t \sqrt{6} = 300 \text{ nm}$ $t = \frac{300}{\sqrt{6}} \text{ nm}$

Rationalizing the denominator: $t = \frac{300\sqrt{6}}{6} \text{ nm} = 50\sqrt{6} \text{ nm}$