Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1004: 20

$4.37cm$

We can determine the width of the central bright fringe as follows: $2y_1=2L tan (sin^{-1}\frac{\lambda}{2d})$ $2y_1=2L tan(sin(\frac{1}{5}sin\theta_3))$ We plug in the known values to obtain: $2y_1=2Ltan[sin^{-1}(\frac{1}{5}tan^{-1}(\frac{0.11m}{1.0m}))]$ $\implies 2y_1=0.0437m=4.37cm$