Question

The \(8^\text{th}\) bright fringe above the point \( O \) oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (\(\mu \text{m}\)), is.

Hint:

To calculate fringe shifts due to oscillating slits, consider both extreme values of slit separation \(d\).

Answer 1

The distance between the slits is given as: \[ d = 0.8 \, \text{mm} \pm 0.04 \, \text{mm}. \] Thus, the extreme values are: \[ d = 0.84 \, \text{mm} \quad \text{and} \quad d = 0.76 \, \text{mm}. \] The fringe position is given by: \[ y = \frac{n \lambda D}{d}, \] where \( n = 8, \lambda = 6 \times 10^{-7} \, \text{m}, D = 1 \, \text{m} \). The separation between the extreme positions is: \[ \Delta y = n \lambda D \left( \frac{1}{d_{\text{min}}} - \frac{1}{d_{\text{max}}} \right). \] Substitute the values: \[ \Delta y = 8 * 6 \times 10^{-7} * 1 \cdot \left( \frac{1}{0.76 \times 10^{-3}} - \frac{1}{0.84 \times 10^{-3}} \right). \] Simplify: \[ \Delta y = 601.50 \, \mu\text{m}. \]