In a young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength $$\lambda = 500$$ nm is incident on the slits. The total number of bright fringes that are observed in the angular range $$-30^{\circ} \leq \theta \leq 30^{\circ}$$ is:

In a Young’s double-slit experiment the condition for bright (constructive) interference is stated first:

$$d \sin\theta = m\lambda$$

Here, $$d$$ is the slit separation, $$\theta$$ is the angular position of the bright fringe with respect to the central axis, $$\lambda$$ is the wavelength of light, and $$m$$ is an integer (… , -3, -2, -1, 0, 1, 2, 3, …).

We are told that

$$d = 0.320\ \text{mm} = 0.320 \times 10^{-3}\ \text{m} = 3.20 \times 10^{-4}\ \text{m},$$

$$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5.00 \times 10^{-7}\ \text{m},$$

and that the observation is restricted to the angular range

$$-30^{\circ} \le \theta \le 30^{\circ}.$$

For any bright fringe to lie inside this range, the following inequality must hold:

$$|\,\sin\theta\,| \le \sin 30^{\circ} = \tfrac12.$$

Using the bright-fringe formula, we rewrite this requirement in terms of $$m$$:

$$\left|\frac{m\lambda}{d}\right| \le \tfrac12.$$

Isolating $$m$$ gives

$$|m| \le \frac{d}{\lambda}\,\frac12.$$

We now compute the numerical factor $$\dfrac{d}{\lambda}$$ first:

$$\frac{d}{\lambda} = \frac{3.20 \times 10^{-4}}{5.00 \times 10^{-7}} = \frac{3.20}{5.00} \times 10^{\,(-4+7)} = 0.64 \times 10^{3} = 640.$$

Multiplying by $$\tfrac12$$ (that is, 0.5) we obtain

$$\frac{d}{\lambda}\,\frac12 = 640 \times 0.5 = 320.$$

Hence the largest integer value (in magnitude) that $$m$$ can take is

$$m_{\text{max}} = 320.$$

Therefore, allowable orders run from $$m = -320$$ up through $$m = +320,$$ inclusive. The total count of integers in this set is calculated as follows:

$$N = (320 - (-320)) + 1 = 640 + 1 = 641.$$

This number $$N$$ is precisely the total number of bright fringes observed within the stated angular limits.

Hence, the correct answer is Option B.