GSO001 Problem 8

This is an application problem combining the change in optical path length with the formula for path difference in Young's experiment.

1. Change in Path Difference

In Young's experiment, the geometric path difference at position $x$ on the screen is approximately $\Delta l \approx \frac{dx}{L}$ .

Placing the film at $S_1$ increases the optical path length through $S_1$ by $(n-1)t$ (the difference between the optical path length $nt$ and the original air path length $1 \cdot t$ ).

2. Condition for the New Central Bright Fringe

At the new central bright fringe, the optical path lengths of the two paths are equal. Since the path through $S_2$ must be geometrically longer, the fringe shifts to $x > 0$ (toward $S_1$ ).

The condition for zero path difference:

\frac{dx_1}{L} - (n-1)t = 0

Solving for $x_1$ :

x_1 = \frac{L(n-1)t}{d}

3. Numerical Calculation

Substituting the constraint values:

x_1 = \frac{1.5 \times (1.4 - 1.0) \times (2.0 \times 10^{-5})}{5.0 \times 10^{-4}}

x_1 = \frac{1.5 \times 0.4 \times 2.0 \times 10^{-5}}{5.0 \times 10^{-4}} = \frac{1.2 \times 10^{-5}}{5.0 \times 10^{-4}}

x_1 = 0.24 \times 10^{-1} = 0.024 \text{ m}

Converting to mm:

x_1 = 0.024 \times 1000 = 24 \text{ mm}

Answer: 24

GSO001 Problem 8

Young's Experiment: Fringe Shift Due to Refractive Index

Problem Statement

Constraints

Input Format

Solution

1. Change in Path Difference

2. Condition for the New Central Bright Fringe

3. Numerical Calculation