Orange light of wavelength $$6000 \times 10^{-10}$$ m illuminates a single slit of width $$0.6 \times 10^{-4}$$ m. The maximum possible number of diffraction minima produced on both sides of the central maximum is __________

We have a single-slit diffraction experiment in which monochromatic light of wavelength $$\lambda = 6000 \times 10^{-10}\,\text{m}$$ is incident on a slit of width $$a = 0.6 \times 10^{-4}\,\text{m}$$.

First we convert these numbers to the same power of ten so that later division is easy.

For the wavelength:

$$\lambda = 6000 \times 10^{-10}\,\text{m} = 6.0 \times 10^{-7}\,\text{m}.$$

For the slit width:

$$a = 0.6 \times 10^{-4}\,\text{m} = 6.0 \times 10^{-5}\,\text{m}.$$

The condition for the minima in a single-slit diffraction pattern is stated by the formula

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

where $$m = \pm 1, \pm 2, \pm 3, \ldots$$ represents the order of the minima.

Because $$\sin\theta$$ can never exceed 1, the largest integer value that $$m$$ can have must satisfy

$$a \sin\theta \le a \quad\Longrightarrow\quad m\lambda \le a,$$

which gives

$$m_{\text{max}} = \frac{a}{\lambda}.$$

Substituting the numerical values, we obtain

$$m_{\text{max}} = \frac{6.0 \times 10^{-5}}{6.0 \times 10^{-7}}.$$

Now we divide the coefficients and subtract the exponents of ten:

$$\frac{6.0}{6.0} = 1,$$

and

$${10^{-5}} \div {10^{-7}} = 10^{-5 - (-7)} = 10^{2}.$$

Hence

$$m_{\text{max}} = 1 \times 10^{2} = 100.$$

This means that on one side of the central maximum we can have integers $$m = 1, 2, 3, \ldots, 100,$$ i.e. exactly $$100$$ minima.

Because the diffraction pattern is symmetric, the same number of minima appear on the other side. Therefore the total number of minima on both sides is

$$2 \times 100 = 200.$$

So, the answer is $$200$$.