Question

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 dark fringe

Answer 1

Answer:

The thickness is  $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

   The wavelength is  $\lambda = 480 \ nm = 480*10^{-9} \ m$

    The first order of the dark  fringe is  $m_1 = 16$

     The second order of dark fringe considered is  $m_2 = 6$

Generally the condition for destructive interference is mathematically represented as

        $y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      $y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=>  $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=>  $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=>  $\Delta y = 2.4 *10^{-6} \ m$