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(B) The fringe width $\beta$ of a double-slit interference pattern is given by the formula: $\beta = \frac{D \lambda}{d}$In this expression,$D$ is the

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decrease the distance between $S_2$ and $S_3$

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(B) The fringe width $\beta$ of a double-slit interference pattern is given by the formula: $\beta = \frac{D \lambda}{d}$In this expression,$D$ is the distance between the screen (or microscope) and the plane containing the slits,$\lambda$ is the wavelength of the light used,and $d$ is the distance between the two coherent sources (slits $S_2$ and $S_3$).In the given setup,the distance $d$ between $S_2$ and $S_3$ is $5 	ext{ cm}$,which is very large for a typical Young's double-slit experiment. Because $\beta$ is inversely proportional to $d$,a large value of $d$ results in extremely narrow fringes that are too close together to be resolved by the microscope $M$.To make the interference fringes visible,the fringe width $\beta$ must be increased. Since $\beta \propto \frac{1}{d}$,the student should decrease the distance $d$ between the slits $S_2$ and $S_3$.

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