In a single slit diffraction pattern, a light of wavelength \( 6000 \, \text{Å} \) is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from the slits. The width of the slit is ______ \( \times 10^{-4} \) m.
Correct Answer: 2
Solution and Explanation
To find the width of the slit in a single slit diffraction pattern, we can use the formula for the position of minima: \(a \sin \theta = n\lambda\), where \(a\) is the slit width, \(\theta\) is the diffraction angle, \(n\) is the order of the minima, and \(\lambda\) is the wavelength of light. For small angles, \(\sin \theta \approx \tan \theta = \frac{y}{D}\), where \(y\) is the distance of the minima from the central maximum on the screen, and \(D\) is the distance from the slit to the screen.
The distance between the first (\(n=1\)) and third (\(n=3\)) minima is given as 3 mm. Using the positions of the minima: \(y_1 = \frac{\lambda D}{a}\) and \(y_3 = \frac{3\lambda D}{a}\).
The difference is: \(y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2\lambda D}{a}\).
Equating this to the given distance: \(\frac{2\lambda D}{a} = 3 \, \text{mm} = 0.003 \, \text{m}\).
Substitute \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m}\) and \(D = 50 \, \text{cm} = 0.5 \, \text{m}\):
\(a = \frac{2 \cdot 6000 \times 10^{-10} \cdot 0.5}{0.003}\).
\(a = \frac{6000 \times 10^{-10}}{0.003} = 2 \times 10^{-4} \, \text{m}\).
Thus, the width of the slit is \(2 \times 10^{-4} \, \text{m}\), fitting well within the specified range [2,2].
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