Question

A thin transparent film of thickness 3000 Å and refractive index 1.5 is layered on sheet of metal. Assuming viewed normally what will be wavelength of visible wavelength observed after reflection from the metal.

• A. 18000 Å
• B. 4500 Å
• C. 5000 Å
• D. 3600 Å

Answer 1

Answer: (B)

4500 Å

Answer 2

The phenomenon observed is thin film interference. Light waves reflected from the top and bottom surfaces of the film interfere. The path difference between the two reflected rays is $2nt$. Given $t = 3000$ Å and $n = 1.5$, the path difference is $2 \times 1.5 \times 3000$ Å $= 9000$ Å.

There is a phase change of $\pi$ at both the air-film interface and the film-metal interface. Since both reflections cause a $\pi$ phase change, the total phase difference due to reflections is $2\pi$.

For constructive interference, the total phase difference must be an integer multiple of $2\pi$: $\frac{2\pi}{\lambda_{vac}}(2nt) + 2\pi = 2m\pi$ $2nt = (m-1)\lambda_{vac}$ Let $m' = m-1$. Then $2nt = m'\lambda_{vac}$.

Substituting the values: $9000$ Å $= m'\lambda_{vac}$. For $m'=2$, $\lambda_{vac} = 9000/2 = 4500$ Å, which is in the visible spectrum.