Question

In a diffraction pattern of a monochromatic light of wavelength 6000 pm, the slit width is 3 mm. If the angular position of 2nd minima is N x 10^-6 rad, find N.

Answer 1

Correct Answer: 4

The diffraction pattern for a single slit can be understood using the formula for the position of minima: \(a \sin \theta = m\lambda\), where \(a\) is the slit width, \(\theta\) is the angular position of the minima, \(m\) is the order of the minima, and \(\lambda\) is the wavelength of the light.

Given: \(\lambda = 6000\) pm (converted to meters: \(6000 \times 10^{-12}\) m), slit width \(a = 3\) mm (converted to meters: \(3 \times 10^{-3}\) m), and we are focusing on the 2nd minima (\(m = 2\)).

We calculate \(\sin \theta\) using:

\[a \sin \theta = m \lambda\]

\[\sin \theta = \frac{m \lambda}{a} = \frac{2 \times 6000 \times 10^{-12}}{3 \times 10^{-3}}\]

\[\sin \theta = \frac{12000 \times 10^{-12}}{3 \times 10^{-3}}\]

\[\sin \theta = 4 \times 10^{-6}\]

For small angles, \(\sin \theta\) approximates \(\theta\), thus

\(\theta \approx 4 \times 10^{-6}\) rad.

Therefore, N = 4, fitting within the specified range [4, 4]. The value of N is precisely 4.