Question

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 \(\dots\) \(\times \, 10^{-4} \, \text{m}\).

Answer 1

Correct Answer: 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:
\[a = 2 \times 10^{-4} \, \text{m}.\]