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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Wavelength of green light, $ \lambda = 5303 \, \mathop{\text{Å}}\limits^{} = 5303 \times 10^{-10} \text{ m} $
• Slit separation, $ d = 19.44 \, \mu \text{m} = 19.44 \times 10^{-6} \text{ m} $
• Slit width, $ a = 4.05 \, \mu \text{m} = 4.05 \times 10^{-6} \text{ m} $
Step 2: Recall the Concept of Interference Envelopes
In a double-slit experiment, the interference pattern (with fringe spacing determined by $ d $) is modulated by the diffraction envelope (determined by $ a $). The single-slit diffraction minima occur when:
$ a \sin \theta = m \lambda \quad (m = 1, 2, 3, \dots) $
The interference maxima occur more frequently, with angular separation:
$ \Delta \theta_{\text{int}} \approx \frac{\lambda}{d} $
Step 3: Determine the Number of Bright Fringes Within One Diffraction Order
Between two successive diffraction minima (for example, between $ m=1 $ and $ m=2 $), the angular spread is approximately
$ \Delta \theta_{\text{diff}} = \frac{\lambda}{a} $.
The number of interference bright fringes (maxima) that fit between two such minima is roughly:
$ \text{Number of fringes} = \frac{\Delta \theta_{\text{diff}}}{\Delta \theta_{\text{int}}} = \frac{\lambda/a}{\lambda/d} = \frac{d}{a} $
Step 4: Substitute the Values
$ \frac{d}{a} = \frac{19.44 \times 10^{-6}}{4.05 \times 10^{-6}} \approx 4.8 $
This result is close to 5, meaning there are 5 interference bright fringes between the first and second diffraction minima.
Step 5: Conclude the Number of Bright Fringes
Hence, the number of bright fringes between the first and second diffraction minima is 5.
