In a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of $$19.44 \mu m$$ and a width of $$4.05 \mu m$$. The number of bright fringes between the first and the second diffraction minima is

We first recall that in a double-slit set-up the intensity pattern observed on the screen is the product of two separate effects:

1. Single-slit diffraction from each slit of width $$a$$ gives dark (minimum-intensity) directions obeying the well-known condition

$$a\,\sin\theta \;=\; m\,\lambda,$$

where $$m = \pm 1, \pm 2, \pm 3,\dots$$. The values of $$m$$ label the successive diffraction minima. The first minimum corresponds to $$m = 1$$, the second to $$m = 2$$, and so on.

2. Interference between the two slits of centre-to-centre separation $$d$$ produces bright fringes in the directions that satisfy

$$d\,\sin\theta \;=\; n\,\lambda,$$

with $$n = 0, \pm 1, \pm 2,\dots$$. Each integer $$n$$ therefore labels one interference maximum (bright fringe).

We are asked to count how many interference maxima lie between the first and the second diffraction minima, i.e. for angles satisfying

$$\text{first minimum:}\; a\sin\theta = 1\,\lambda \quad \text{to} \quad \text{second minimum:}\; a\sin\theta = 2\,\lambda.$$

To proceed, we substitute the expression for $$\sin\theta$$ obtained from the diffraction condition into the interference condition so that the two formulae can be linked directly.

From the diffraction minima condition we have

$$\sin\theta = \dfrac{m\,\lambda}{a}.$$

Placing this value of $$\sin\theta$$ into the interference condition $$d\sin\theta = n\lambda$$ gives

$$d \left(\dfrac{m\,\lambda}{a}\right) = n\,\lambda.$$

The factor $$\lambda$$ cancels on both sides, leaving

$$n = \dfrac{d}{a}\,m.$$

This remarkably simple relation tells us that whenever $$m$$ takes an integer value at a diffraction minimum, the corresponding interference order is

$$n = \left(\dfrac{d}{a}\right)m.$$

We now put in the numerical values supplied in the question:

$$d = 19.44\;\mu \text{m}, \quad a = 4.05\;\mu \text{m}.$$

Hence

$$\dfrac{d}{a} = \dfrac{19.44}{4.05} = 4.8.$$

Let us evaluate $$n$$ at the two diffraction minima in question.

For the first diffraction minimum ($$m = 1$$) we obtain

$$n_1 = (4.8)(1) = 4.8.$$

For the second diffraction minimum ($$m = 2$$) we obtain

$$n_2 = (4.8)(2) = 9.6.$$

Between these two values of $$\theta$$, the interference maxima correspond to integer $$n$$ lying strictly between 4.8 and 9.6 (because at the minima themselves the intensity is zero, so those angular positions are excluded). All integers satisfying

$$4.8 < n < 9.6$$

are therefore counted. Listing them explicitly, we have

$$n = 5,\;6,\;7,\;8,\;9.$$

This is a total of

$$5 \text{ bright fringes}.$$

Hence, the correct answer is Option B.