1. Condition for minima in single-slit diffraction

For a single slit of width $a$, diffraction minima occur at

$a\sin\theta = m\lambda$

where $m = 1,2,3,\dots$

2. Maximum possible order of minima

Since $\sin\theta \le 1$, the maximum possible value of $m$ satisfies

$m\lambda \le a$

So,

$m_{\max} = \left\lfloor \frac{a}{\lambda} \right\rfloor$

Given:

$\lambda = 6000\times 10^{-10}\,\text{m} = 6\times 10^{-7}\,\text{m}$

$a = 0.6\times 10^{-4}\,\text{m} = 6\times 10^{-5}\,\text{m}$

Thus,

$\frac{a}{\lambda} = \frac{6\times 10^{-5}}{6\times 10^{-7}} = 10^2 = 100$

Hence,

$m_{\max} = 100$

3. Total number of minima on both sides

For each order $m = 1$ to $100$, there is one minimum on the left and one on the right of the central maximum.

Therefore, total number of minima on both sides is

$N = 2\times 100 = 200$

4. Final answer

$\boxed{200}$

5. Comparison with stored answer

Stored correct answer = $200$

Our derived answer matches the stored answer.