Answer:
0.00915 m
Explanation:
[tex]\lambda[/tex] = Wavelength of light = 610 nm
L = Distance of slit to screen = 1.5 m
a = Slit gap = 0.2 mm
Width of central maximum is given by
[tex]w=\dfrac{2\lambda L}{a}\\\Rightarrow w=\dfrac{2\times 610\times 10^{-9}\times 1.5}{0.2\times 10^{-3}}\\\Rightarrow w=0.00915\ m[/tex]
The width of the central maximum is 0.00915 m
