Diffraction: Practice Problem & Solution
A screen is placed 50 cm from a single slit, which is illuminated with 6000 Å light. If distance between the first and third minima in the diffraction patten is 3 mm, the width of the slit is
Solution Explained:
To solve this problem, we apply the core principles of Diffraction. Understanding the underlying formula is key to arriving at the correct answer below:
The position of \(n^{\text{th}}\) minimum in single-slit diffraction is \(y_n = \frac{n\lambda D}{a}\). The distance between the 1st and 3rd minima is \(Delta y = \frac{2\lambda D}{a} ⇒ a = \frac{2\lambda D}{\Delta y} = \frac{2 \times 6000 \times 10^{-10} \times 0.50}{3 \times 10^{-3}} = 0.2\text{ mm}\).
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