In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is $$x \times 10^{-3}$$ cm, where x-value is ______.

For a single slit of width $$a$$, the first diffraction minima occur at angles $$\theta$$ that satisfy

$$a \sin \theta = \pm \lambda \qquad -(1)$$

Hence the central diffraction maximum extends from $$-\theta_1$$ to $$+\theta_1$$ where

$$\sin \theta_1 = \frac{\lambda}{a} \qquad -(2)$$

For the Young’s double-slit interference, bright fringes (maxima) are obtained at

$$d \sin \theta = m \lambda ,\; m = 0, \pm 1, \pm 2, \ldots \qquad -(3)$$

The extreme interference maximum that can still lie inside the central diffraction maximum is obtained by equating the conditions $$(2)$$ and $$(3)$$:

$$d \sin \theta_1 = m_{\text{max}}\lambda$$
$$\Rightarrow d \left(\frac{\lambda}{a}\right) = m_{\text{max}}\lambda$$
$$\Rightarrow m_{\text{max}} = \frac{d}{a} \qquad -(4)$$

The problem states that a total of 20 interference maxima are contained inside the central diffraction maximum. These 20 maxima are the bright fringes on both sides of the central bright fringe but excluding the central one itself. Therefore

Number of maxima on one side $$= \frac{20}{2}=10$$
$$\Rightarrow m_{\text{max}} = 10 \qquad -(5)$$

Substituting $$(5)$$ into $$(4)$$ gives the required slit width:

$$a = \frac{d}{m_{\text{max}}} = \frac{1.5 \text{ mm}}{10} = 0.15 \text{ mm}$$

Converting to centimetres,

$$0.15 \text{ mm} = 0.015 \text{ cm} = 15 \times 10^{-3} \text{ cm}$$

Thus, the width of each slit is $$15 \times 10^{-3}$$ cm, so the required value of $$x$$ is $$15$$.