Step 1: Determine the Phase Difference

Let the screen lie along the $x$-axis. We consider the phase of the two plane waves at a position $x$ on the screen.

• Wave 1 is incident at angle $\theta_1$. Its wavefronts define a path difference relative to the origin of $\Delta x_1 = x \sin \theta_1$.
• Wave 2 is incident at angle $\theta_2$. Its path difference is $\Delta x_2 = x \sin \theta_2$.

The net path difference $\Delta$ between the two waves at position $x$ is:

$$ \Delta = x(\sin \theta_2 – \sin \theta_1) $$

Step 2: Condition for Adjacent Fringes

Interference maxima (bright fringes) occur when the path difference is an integer multiple of the wavelength $\lambda$:

$$ \Delta = n\lambda \implies x(\sin \theta_2 – \sin \theta_1) = n\lambda $$

To find the fringe width $w$ (the distance between adjacent maxima), we look for the change in $x$ that corresponds to a change in path difference of exactly one wavelength ($\lambda$).

$$ w (\sin \theta_2 – \sin \theta_1) = \lambda $$

Step 3: Solve for Fringe Width

Rearranging the equation for $w$:

$$ w = \frac{\lambda}{\sin \theta_2 – \sin \theta_1} $$

Note: Since the direction of angles affects the sign, generally the denominator is the magnitude of the difference in sine components, $|\sin \theta_2 – \sin \theta_1|$. Based on the diagram where angles are on opposite sides of the normal, the effective angular separation contributes constructively to the phase gradient.