Question

Obtain the formula for the distance of $n^{th}$ order dark fringe from the central fringe with the help of Young’s double-slit experiment.

Hint:

In Young’s double slit experiment: - Bright fringe: $y_n = n \frac{\lambda D}{d}$ - Dark fringe: $y_n = (n - \tfrac{1}{2}) \frac{\lambda D}{d}$

Answer 1

Step 1: Recall condition for interference.
In Young’s double slit experiment, interference fringes are formed due to path difference between the two waves from slits.
Step 2: Condition for dark fringes.
Dark fringe occurs when: \[ \Delta = \left( n - \tfrac{1}{2} \right) \lambda, \quad n = 1, 2, 3 \dots \]
Step 3: Relation between path difference and fringe position.
Path difference at distance $y$ from central maximum is: \[ \Delta = \frac{y d}{D}, \] where $d$ = distance between slits, $D$ = distance between slits and screen.
Step 4: Substitute condition.
\[ \frac{y_n d}{D} = \left( n - \tfrac{1}{2} \right) \lambda. \]
Step 5: Solve for $y_n$.
\[ y_n = \left( n - \tfrac{1}{2} \right) \frac{\lambda D}{d}. \]
Step 6: Conclusion.
Thus, the distance of the $n^{th}$ dark fringe from the central fringe is: \[ y_n = \left( n - \tfrac{1}{2} \right) \frac{\lambda D}{d}. \]