Question

For a two-slit Fraunhofer diffraction, each slit is 0.1 mm wide and separation between the two slits is 0.8 mm. The total number of interference minima between the first diffraction minima on both sides of the central maxima is

• A. 16
• B. 18
• C. 8
• D. 9

Hint:

A useful rule of thumb for these problems is to calculate the ratio \(d/a\). The total number of interference maxima within the central diffraction maximum is \(2(d/a) - 1\) (if \(d/a\) is an integer), and the number of interference minima is \(2(d/a)\). Here, \(d/a = 8\), so the number of minima is \(2 \times 8 = 16\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a two-slit diffraction pattern, we observe rapid interference fringes modulated by a slower diffraction envelope. The question asks for the number of interference minima that lie within the central diffraction maximum. The central diffraction maximum is the bright region between the first diffraction minimum on the left and the first diffraction minimum on the right.
Step 2: Key Formula or Approach:
Let \(a\) be the slit width and \(d\) be the separation between the slits.

The condition for diffraction minima is given by: \(a \sin\theta = n\lambda\), for \(n = \pm 1, \pm 2, \dots\)
The condition for interference minima is given by: \(d \sin\theta = (m + \frac{1}{2})\lambda\), for \(m = 0, \pm 1, \pm 2, \dots\)
We need to find the number of interference minima whose angles \(\theta\) are smaller in magnitude than the angle of the first diffraction minimum.
Step 3: Detailed Explanation:
1. Find the boundary of the central diffraction maximum: The central diffraction maximum is bounded by the first diffraction minima (\(n=1\)). Let the angle for the first diffraction minimum be \(\theta_D\). \[ a \sin\theta_D = 1 \cdot \lambda \implies \sin\theta_D = \frac{\lambda}{a} \] So, the central maximum extends over the angular range \(-\frac{\lambda}{a}<\sin\theta<\frac{\lambda}{a}\). 2. Find the positions of interference minima: Let the angle for an interference minimum be \(\theta_I\). \[ d \sin\theta_I = (m + \frac{1}{2})\lambda \implies \sin\theta_I = (m + \frac{1}{2})\frac{\lambda}{d} \] 3. Apply the condition: We need to find the number of interference minima that lie within the central diffraction maximum. This means we need to find the integer values of \(m\) that satisfy: \[ |\sin\theta_I|<|\sin\theta_D| \] \[ \left|(m + \frac{1}{2})\frac{\lambda}{d}\right|<\left|\frac{\lambda}{a}\right| \] The \(\lambda\) cancels out: \[ \left|m + \frac{1}{2}\right|<\frac{d}{a} \] 4. Substitute the given values: Slit width \(a = 0.1\) mm. Slit separation \(d = 0.8\) mm. \[ \frac{d}{a} = \frac{0.8}{0.1} = 8 \] The inequality becomes: \[ \left|m + \frac{1}{2}\right|<8 \] This can be written as: \[ -8<m + \frac{1}{2}<8 \] Subtract \(\frac{1}{2}\) from all parts: \[ -8.5<m<7.5 \] 5. Count the number of integer values for m: The possible integer values for \(m\) are: \(-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7\). The total number of values is the number of integers from -8 to 7, inclusive. Number of values = (Last) - (First) + 1 = \(7 - (-8) + 1 = 7 + 8 + 1 = 16\). So, there are a total of 16 interference minima within the central diffraction maximum. Step 4: Final Answer:
The total number of interference minima between the first diffraction minima on both sides of the central maxima is 16. This corresponds to option (A).