Solution
Given,
Distance between the center of the interference pattern and the tenth bright fringe (y) = 3.44 cm = 3.44×10-2 m,
Distance between slits and screen (D) = 2m,
The wavelength of light used (λ) = 5.89×10-7 m,
Slit separation (d) =?,
The angle subtended by the central bright fringe (α) =?
We have the relation,
\(y=\frac{n\lambda D}d\)
Or, \(d=\frac{n\lambda D}y\)
Or, \(d=\frac{10\times5.89\times10^{-7}\times2}{0.0344}\)
∴ d = 3.42×10-4 m
Also, we have,
\(\alpha=\frac\beta D=\frac{\lambda D}d\) \(\left[\because\beta=\frac{\lambda D}d\right]\)
Or, \(\alpha=\frac\lambda d=\frac{5.89\times10^{-7}}{3.42\times10^{-4}}\)
∴ α = 1.72×10-3 radians
Hence, the separation between slit is 3.42×10^-4 m and the angle made is 1.72×10^-3 radians.