Question

Maxima of light of wavelength $\lambda$ is observed in thin film interference for reflection. Now the thickness of film is slowly increased by $\Delta x = 1$ nm such that wavelength is observed to be absent for the first time in reflection spectra. Refractive index of film 1.5 then wavelength of light is N nm then N is

Answer 1

6

Answer 2

The condition for constructive interference (maxima) in reflected light from a thin film, considering a phase change of $\pi$ at one interface, is given by: $2 \mu t \cos r = (m + 1/2) \lambda$ where $\mu$ is the refractive index of the film, $t$ is its thickness, $r$ is the angle of refraction, $\lambda$ is the wavelength of light, and $m$ is an integer ($m = 0, 1, 2, ...$).

The condition for destructive interference (minima) is: $2 \mu t \cos r = m' \lambda$ where $m'$ is an integer ($m' = 1, 2, 3, ...$).

Assuming normal incidence, $\cos r = 1$. The equations become: Maxima: $2 \mu t = (m + 1/2) \lambda$ Minima: $2 \mu t = m' \lambda$

Let the initial thickness be $t$. A maximum is observed for wavelength $\lambda$. $2 \mu t = (m + 1/2) \lambda$ for some integer $m \ge 0$.

The thickness is increased by $\Delta x = 1$ nm, so the new thickness is $t' = t + \Delta x$. For this new thickness, the wavelength $\lambda$ is absent for the first time, meaning it corresponds to the first minimum encountered after the initial maximum. The sequence of conditions for $2\mu t/\lambda$ as thickness increases is: Maxima: $0.5, 1.5, 2.5, 3.5, \ldots$ (corresponding to $m=0, 1, 2, 3, \ldots$) Minima: $1, 2, 3, 4, \ldots$ (corresponding to $m'=1, 2, 3, 4, \ldots$)

If the initial maximum is of order $m$, the very next condition is a minimum of order $m' = m+1$. So, for the new thickness $t'$: $2 \mu (t + \Delta x) = (m+1) \lambda$

Subtracting the initial condition from the new condition: $2 \mu (t + \Delta x) - 2 \mu t = (m+1) \lambda - (m + 1/2) \lambda$ $2 \mu \Delta x = (m+1 - m - 1/2) \lambda$ $2 \mu \Delta x = (1 - 1/2) \lambda$ $2 \mu \Delta x = \frac{1}{2} \lambda$

We are given: Refractive index of film, $\mu = 1.5$ Increase in thickness, $\Delta x = 1$ nm The original wavelength, $\lambda = N$ nm

Substitute these values into the derived equation: $2 \times (1.5) \times (1 \text{ nm}) = \frac{1}{2} \times (N \text{ nm})$ $3 \text{ nm} = \frac{N}{2} \text{ nm}$ $N = 3 \times 2$ $N = 6$

Thus, the wavelength of light is 6 nm.