Question

A physics teacher wants to demonstrate interference with the help of double slit experiment using a laser beam of 633 nm wavelength. Since the hall is large enough, interference pattern is formed on the wall 5.0 m from the slits. For clear and comfortable view by all the students they want the fringe width 5 mm.
(I) Find the slit separation for obtaining the desired interference pattern.
(II) How far will the first minimum be from the central maximum ?

Hint:

Use \(d = \lambda D/\beta\) for (I) and \(y = \beta/2\) for the first minimum in (II).

Answer 1

Step 1: Note the given data:
\(\lambda = 633\ \text{nm} = 6.33\times10^{-7}\ \text{m}\), \(D = 5.0\ \text{m}\) and fringe width \(\beta = 5\ \text{mm} = 5\times10^{-3}\ \text{m}\).

Step 2: Part (I): slit separation:
The fringe width in a double slit experiment is \(\beta = \dfrac{\lambda D}{d}\), so \(d = \dfrac{\lambda D}{\beta}\).
\[ d = \frac{6.33\times10^{-7}\times 5.0}{5\times10^{-3}} = 6.33\times10^{-4}\ \text{m} \]
So the slit separation is \(0.633\ \text{mm}\).

Step 3: Part (II): first minimum:
Dark fringes lie midway between bright fringes. The \(m\)-th minimum is at \(y = \left(m - \tfrac{1}{2}\right)\beta\).
For the first minimum \(m = 1\), so \(y = \dfrac{\beta}{2}\).
\[ y = \frac{5\ \text{mm}}{2} = 2.5\ \text{mm} \]

Final Answer:
The slit separation is \(0.633\) mm and the first minimum is \(2.5\) mm from the central maximum. \[ \boxed{d = 0.633\ \text{mm},\quad y_1 = 2.5\ \text{mm}} \]