Question

Light passes through two media with refractive indices of 1.75 and 1.55, respectively. The thickness of both the media is 30 mm. The resultant path difference of the yellow light component \( \lambda = 589 \, \text{nm} \) is \(\underline{\hspace{1cm}}\) mm.

Hint:

The path difference can be calculated by considering the refractive indices and thickness of the media, ensuring that the refractive indices are taken in the correct order.

Answer 1

Correct Answer: 64

The path difference for light passing through two media is given by:
\[ \Delta x = t \left( \frac{1}{n_1} - \frac{1}{n_2} \right) \] where:
- \( t \) is the thickness of the media,
- \( n_1 \) and \( n_2 \) are the refractive indices of the two media.
Given:
- \( t = 30 \, \text{mm} \),
- \( n_1 = 1.75 \),
- \( n_2 = 1.55 \).
Substituting the values: \[ \Delta x = 30 \left( \frac{1}{1.75} - \frac{1}{1.55} \right) \] \[ \Delta x = 30 \times (0.5714 - 0.6452) = 30 \times (-0.0738) = -2.22 \, \text{mm} \] The resultant path difference is approximately 2.0 mm.