Superposition: Question 8

Worked solution

Step 1: Recall the position of the nth-order bright fringe

The basic double-slit relation $\lambda = \dfrac{ax}{D}$ gives the spacing $x$ between adjacent bright fringes (i.e. the distance from order $n$ to order $n+1$). The distance from the central (zero-order) fringe to the $n$th-order bright fringe is therefore $n$ times this spacing: $x_n = \frac{n\lambda D}{a}$

Step 2: Substitute the given values

Convert to consistent SI units: $\lambda = 500\text{ nm} = 5.00\times10^{-7}\text{ m}$, $a = 0.40\text{ mm} = 4.0\times10^{-4}\text{ m}$, $D = 2.00\text{ m}$, and the required order is $n=3$: $x_3 = \frac{3 \times 5.00\times10^{-7} \times 2.00}{4.0\times10^{-4}}$

Compute the numerator first: $3 \times 5.00\times10^{-7} = 1.50\times10^{-6}$, then $1.50\times10^{-6} \times 2.00 = 3.00\times10^{-6}$.

$x_3 = \frac{3.00\times10^{-6}}{4.0\times10^{-4}} = 7.5\times10^{-3}\text{ m} = 7.5\text{ mm}$

Check (recompute independently): the spacing between adjacent fringes is $x_1 = \lambda D/a = (5.00\times10^{-7}\times2.00)/(4.0\times10^{-4}) = 2.5\times10^{-3}\text{ m} = 2.5\text{ mm}$; three fringe spacings from the centre gives $x_3 = 3\times2.5=7.5\text{ mm}$, the same result.

Step 3: Why the other options are wrong

• A ($2.5$ mm): this is the fringe spacing $x_1$ (order $1$), forgetting to scale up to order $3$.
• B ($5.0$ mm): this uses $n=2$ instead of $n=3$.
• D ($10.0$ mm): this uses $n=4$ instead of $n=3$.

Final answer

• Distance to the third-order bright fringe $= \boxed{7.5\text{ mm}}$, option C.