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Step-by-Step Solution
Step 1: Write down the diffraction minima condition
For a single-slit diffraction pattern, the position of the minima is given by
d \,\sin\theta = n\lambda ,
where
d is the slit width,
λ is the wavelength of the light, and
n is the order of the minima (n = ±1, ±2, ±3, …).
Step 2: Understand the maximum limit of n
The sine of any angle cannot exceed 1. Thus,
\sin\theta \le 1 .
From the diffraction condition,
d \,\sin\theta = n\lambda \implies \sin\theta = \frac{n\lambda}{d} .
Hence,
\frac{n\lambda}{d} \le 1 \implies n \le \frac{d}{\lambda}.
Step 3: Substitute the given values
Given:
d = 0.6 \times 10^{-4}\,\text{m}, \quad \lambda = 6000 \times 10^{-10}\,\text{m}.
So,
n \le \frac{0.6 \times 10^{-4}}{6000 \times 10^{-10}}.
Step 4: Simplify the expression
n \le \frac{0.6 \times 10^{-4}}{6000 \times 10^{-10}}
= \frac{0.6 \times 10^{-4}}{6 \times 10^{-7}}
= \frac{0.6}{6} \times \frac{10^{-4}}{10^{-7}}
= 0.1 \times 10^{3} = 100.
Therefore, the maximum integral value for n is 100 on each side of the central maximum.
Step 5: Calculate the total number of minima
Since there are 100 minima on one side and another 100 on the other side, the total
number of minima on both sides is
100 + 100 = 200.
Final Answer
200
