In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 480 nm, how much thicker is the air wedge at the 16th dark fringe than it is at the 6th?
A) 2400 nm
B) 4800 nm
C) 240 nm
D) 480 nm
E) none of these

Question

Quizplus · Accepted Answer

The thickness difference between two dark fringes in a thin film interference pattern is given by the formula: $ \Delta t = \frac{m \lambda}{2} $, where $ m $ is the fringe order difference. For the 16th and 6th dark fringes, $ m = 16 - 6 = 10 $. Thus, $ \Delta t = \frac{10 \times 480 \, \text{nm}}{2} = 2400 \, \text{nm} $.

In a Thin Film Experiment, a Wedge of Air Is