Physics - Wave Optics Question with Solution | TestHub

Given fringe width $W=0.6\text{ mm}$.

The distance of the $n^{\text{th}}$ bright fringe from the central bright fringe is given by $y_n=nW$.

For the $8^{\text{th}}$ bright fringe, $y_8=8W$.

The distance of the $m^{\text{th}}$ dark fringe from the central bright fringe is given by $y_m^{\prime }=(m-0.5)W$.

For the $6^{\text{th}}$ dark fringe, $y_6^{\prime }=(6-0.5)W=5.5W$.

The distance between the $6^{\text{th}}$ dark fringe and the $8^{\text{th}}$ bright fringe on the same side is:

$\Delta y=y_8-y_6^{\prime }=8W-5.5W=2.5W$

Substituting the value of $W$:

$\Delta y=2.5\times 0.6=1.5\text{ mm}$

Answer: 1.5 mm