Question

Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light wavelength 500 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 ……….. × \( 10^{-4}\) m.

Answer 1

Correct answer is : 2

\(\mathrm{d}=1 \mathrm{~mm}, \mathrm{D}=1 \mathrm{m}, \lambda=500 \mathrm{~nm}\)

\(10\left(\frac{\lambda \mathrm{D}}{\mathrm{d}}\right)=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}\)

\(\mathbf{a}=\frac{\mathrm{d}}{5}\)

\(=\frac{10 \times 10^{-4} \mathrm{~m}}{5}\)

\(=2 \times 10^{-4}\)