Question

A thin film of alcohol is spread over a surface. When light from a tunable source is incident normally, the intensity of reflected light at the detector is maximum for \( \lambda = 640 \, \text{nm} \) and minimum for \( \lambda = 512 \, \text{nm} \. Taking the refractive index of alcohol to be \( 1.36 \) for both the given wavelengths, the minimum thickness of the film would be .......... nm (Round off to two decimal places).

Hint:

For thin films, use the condition for constructive and destructive interference based on the refractive index and wavelength to find the film thickness.

Answer 1

Correct Answer: 470.56

Step 1: Thin film interference.
In thin film interference, the condition for maximum reflection is given by \[ 2 n d = m \lambda \quad \text{(for maxima, where \( m \) is an integer)}. \] For minimum reflection, the condition is \[ 2 n d = (m + \frac{1}{2}) \lambda. \] Step 2: Minimum thickness calculation.
We use the wavelength for minimum intensity \( \lambda = 512 \, \text{nm} \) and the refractive index \( n = 1.36 \) to calculate the minimum thickness of the film. We substitute into the equation for minimum reflection with \( m = 1 \): \[ 2 \times 1.36 \times d = (1 + \frac{1}{2}) \times 512 \, \text{nm}. \] Simplifying, we find \[ 2.72 d = 768 \Rightarrow d = \frac{768}{2.72} = 282.35 \, \text{nm}. \] Final Answer: The minimum thickness of the film is \( \boxed{282.35} \, \text{nm}. \)