Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light wavelength \( 500 \, \text{nm} \) is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is \(\dots\) \(\times \, 10^{-4} \, \text{m}\).
To find the width of each slit that results in 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, we begin by analyzing the interference and diffraction conditions.
For a double-slit, the condition for maxima is: \(d \sin \theta = m\lambda\), where \(d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\) (distance between slits), \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\), and \(m\) is the order of the maximum.
The angular width of the central maximum of single-slit diffraction is given by: \(\sin \theta = \frac{\lambda}{a}\), where \(a\) is the slit width.
Given that 10 maxima of the double-slit pattern fit within the central maximum of the single-slit pattern, the angle for the 5th order (half of 10) double-slit maximum should equal the angle for the first minimum of the single-slit pattern: \(\frac{5\lambda}{d} = \frac{\lambda}{a}\).
This calculated width of the slit \(a = 2 \times 10^{-4} \, \text{m}\) falls within the specified range. Therefore, the width of each slit is \(2 \times 10^{-4} \, \text{m}\).
Was this answer helpful?
0
0
Hide Solution
Verified By Collegedunia
Approach Solution -2
Given: \[d = 1 \, \text{mm} = 10^{-3} \, \text{m}, \quad D = 1 \, \text{m}, \quad \lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}.\] For the central maximum of the single-slit pattern to contain 10 maxima of the double-slit pattern: \[10 \times \frac{\lambda D}{d} = \frac{2\lambda D}{a},\] where $a$ is the slit width. Rearranging for $a$: \[a = \frac{d}{5}.\] Substitute $d = 10^{-3} \, \text{m}$: \[a = \frac{10^{-3}}{5} = 2 \times 10^{-4} \, \text{m}.\] Thus, the slit width is: \[a = 2 \times 10^{-4} \, \text{m}.\]
