Question

The beam of light consisting of two wavelengths $6500\mathring{A}$ and $5200\mathring{A}$ is used to obtain interference fringes in a Young's double slit experiment. The distance between the slits is $2\text{mm}$ and the distance between the plane of the slits and the screen is $120\text{cm}$. What is the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide?

• A. $0.52\text{cm}$
• B. $0.32\text{cm}$
• C. $0.16\text{cm}$
• D. $0.96\text{cm}$

Answer 1

The correct option is C $0.16\text{cm}$

The $n^{\text{th}}$ bright fringe is at a distance $y=\frac{n\lambda D}{d}$ from the central maximum.

Suppose $x^{\text{th}}$ and $y^{\text{th}}$ bright fringe of $6500\mathring{A}$ and $5200\mathring{A}$ coincide respectively, then,
$\frac{x\times 6500\times D}{d}=\frac{y\times 5200\times D}{d}$

$\Rightarrow \frac{x}{y}=\frac{5200}{6500}=\frac{4}{5}$

The minimum integeral value of $x$ and $y$, that satisfies this equation are $4$ and $5$ respectively.

So, the distance of $5^{\text{th}}$ bright fringe, due to the wavelength of $6500\mathring{A}$, from central maximum is,

$y=\frac{4\times 6500\times {10}^{-10}\times 1.2}{2\times {10}^{-3}}=0.156\text{cm}$

$\therefore y\approx 0.16\text{cm}$

Hence, $\left(C\right)$ is the correct answer.

Why this question? Learning to figure out common points about interference pattern due to two different wavelengths.