second-order-young-fringe-doubles-position

Because fringe position is linear in the order m, the second-order bright fringe lands twice as far from the center as the first. Doubling m simply doubles y in the rule y_m = m * lambda * L / d. The exact doubling holds under the same ideal conditions: coherent monochromatic light, small angles, and known geometry; real fringes shift with alignment error, slit-width effects, and screen calibration.

The second bright source doubles the screen position with the same geometry. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Second order needs a second-order path source

The source path difference is 4 m, exactly two wavelengths. That source order must match the screen order.

\Delta=4\ \text{m}=2\lambda

The second row is twice the first

With the same geometry, the second row lands at twice the first screen position. The third row keeps the same scale visible.

\begin{array}{c|c|c|c|c}m&\lambda&L&d&y\\1&2\ \text{m}&6\ \text{m}&3\ \text{m}&4\ \text{m}\\2&2\ \text{m}&6\ \text{m}&3\ \text{m}&8\ \text{m}\\3&2\ \text{m}&6\ \text{m}&3\ \text{m}&12\ \text{m}\\\end{array}

Doubling the order doubles the position

The second-order source has path difference 4 m, so the screen position doubles with the same slit and screen distance.

y=2\frac{\lambda L}{d}=8\ \text{m};\quad \Delta=4\ \text{m}