Question

In a double-slit experiment, the slit separation d is 2.00 times the slit width w. How many bright interference fringes are in the central diffraction envelope?

Short answer

The number of bright interference fringes that are in the central diffraction envelope is 3.

Step-by-step solution

Concept/Significance of double slit experiment

For the first minima in the diffraction pattern,

$\mathit{w}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}{\mathit{m}}_{\mathbf{1}}\mathit{\lambda }$

The angular locations of the bright fringes of the double-slit interference pattern is given by,

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}{\mathit{m}}_{\mathbf{2}}\mathit{\lambda }$ …… (1)

Here, d is the slit separation, and w is single slit width.

Find the number of bright interference fringes that are in the central diffraction envelope

For the 1^st minima $m_1=1$, then rewrite the equation (1) as follows.

$a\sin \theta =\lambda$ …… (2)

From the equations (1) and (2),

$$\begin{gathered} m_2=\frac{d}{w} \\ =\frac{2w}{w} \\ =2.0 \end{gathered}$$

This means that inside central diffraction envelope there are three bright fringes, two in each side and one at the centre.

Therefore, the number of bright interference fringes that are in the central diffraction envelope is 3 .