In a Youngs double slit experiment, the distance between the slits is $d=0.1\text{ mm}$ and the screen is $D=1\text{ m}$ away. If the wavelength of light used is $\lambda =500\text{ nm}$, what is the distance between the third bright fringe and the central maximum?

Solution

1. Identify the given values: slit separation $$\begin{array}{rl}d& =0.1\text{ mm}\\ & =10^{-4}\text{ m}\end{array}$$, screen distance $D=1\text{ m}$, wavelength $$\begin{array}{rl}\lambda & =500\text{ nm}\\ & =5\times 10^{-7}\text{ m}\end{array}$$, and fringe order $n=3$.
2. Use the formula for the position of the $n^{\text{th}}$ bright fringe: $y_n=\frac{n\lambda D}{d}$.
3. Substitute the values into the equation: $y_3=\frac{3\times (5\times 10^{-7}\text{ m})\times 1\text{ m}}{10^{-4}\text{ m}}$.
4. Calculate the result: $$\begin{array}{rl}{y}_3& =15\times 10^{-3}\text{ m}\\ & =15\text{ mm}\end{array}$$.
5. Hence the answer is (B).