Single slit diffraction pattern — derive width of central maximum

Question

Light of wavelength 600 nm passes through a single slit of width 0.1 mm. The screen is 1 m away. Derive the condition for the first minimum and find the width of the central maximum.

(JEE Advanced 2023, similar pattern)

Solution — Step by Step

Consider a single slit of width $a$ . We divide the slit into two equal halves. For each point in the upper half, there is a corresponding point in the lower half at distance $a/2$ .

At the first minimum, the path difference between these paired points equals $\lambda/2$ . This gives:

a \sin\theta = \lambda \quad \text{(first minimum)}

In general, the $n^\text{th}$ minimum occurs at $a\sin\theta = n\lambda$ .

\sin\theta_1 = \frac{\lambda}{a} = \frac{600 \times 10^{-9}}{0.1 \times 10^{-3}} = 6 \times 10^{-3}

Since $\theta$ is very small: $\theta_1 \approx \sin\theta_1 = 6 \times 10^{-3}$ rad.

The first minimum appears at distance $y_1$ from the centre:

y_1 = D\tan\theta_1 \approx D\theta_1 = 1 \times 6 \times 10^{-3} = 6 \times 10^{-3} \text{ m} = 6 \text{ mm}

The central maximum extends from the first minimum on one side to the first minimum on the other side:

\text{Width} = 2y_1 = 2 \times 6 = \mathbf{12 \text{ mm}}

Using the formula directly: Width of central maximum $= \dfrac{2\lambda D}{a}$ .

Why This Works

Single slit diffraction occurs because different parts of the slit act as coherent sources (Huygens’ principle). At the centre ( $\theta = 0$ ), all wavelets arrive in phase — constructive interference gives the bright central maximum.

At the first minimum, we pair up sources from the two halves of the slit. Each pair has a path difference of $\lambda/2$ , so each pair destructively interferes. When ALL pairs cancel, we get zero intensity — the first dark fringe.

The central maximum is twice as wide as the other maxima. The secondary maxima get progressively dimmer — the first secondary maximum has only about 4.5% of the central maximum’s intensity.

Alternative Method

Use the intensity formula: $I(\theta) = I_0 \left(\frac{\sin\beta}{\beta}\right)^2$ where $\beta = \frac{\pi a \sin\theta}{\lambda}$ . The first minimum occurs when $\sin\beta = 0$ and $\beta \neq 0$ , i.e., $\beta = \pi$ , giving $a\sin\theta = \lambda$ . Same condition.

Compare with Young’s double slit: in YDSE, fringe width $= \lambda D/d$ and all fringes are equally spaced. In single slit, the central maximum width $= 2\lambda D/a$ (twice the secondary fringe width). JEE loves asking you to compare these two patterns.

Common Mistake

Students confuse the condition for minima in single slit ( $a\sin\theta = n\lambda$ ) with the condition for maxima in double slit ( $d\sin\theta = n\lambda$ ). Same formula structure, but one gives dark fringes and the other gives bright fringes. The mnemonic: Single slit, Same formula = dark (minima). Double slit, Same formula = bright (maxima).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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