Answer:
The width is
![w_c = 0.00422 \ m](https://img.qammunity.org/2021/formulas/physics/college/bufil9qvkk46dy80lbi2h8sob3u8jmegcp.png)
Step-by-step explanation:
From the question we are told that
The wavelength is
![\lambda = 6.33*10^(-7) \ m](https://img.qammunity.org/2021/formulas/physics/college/pwz5fdruqzqcx792tp69km0m948mgdw3pw.png)
The width of the slit is
![d = 0.3\ mm = 0.3 *10^(-3) \ m](https://img.qammunity.org/2021/formulas/physics/college/9xqzuy2brswykjqsj47x4qbudigprso0sq.png)
The distance of the screen is
![D = 1.0 \ m](https://img.qammunity.org/2021/formulas/physics/college/6pnjvx45x6z25z354f24b1i3wkexk7gjps.png)
Generally the central maximum is mathematically represented as
![w_c = 2 * y](https://img.qammunity.org/2021/formulas/physics/college/bsbkjguymvkjpia0qzo46ebv8l3ctbtsel.png)
Here y is the width of the first order maxima which is mathematically represented as
![y = (\lambda * D)/(d)](https://img.qammunity.org/2021/formulas/physics/college/11ezexpz17jwp61ma8izku0o0i562vsqxt.png)
substituting values
![y = (6.33*10^(-7) * 1.0)/( 0.30)](https://img.qammunity.org/2021/formulas/physics/college/6x2fkkf61rkw0v8ajsv9vy2zhsludn7pad.png)
![y = 0.00211 \ m](https://img.qammunity.org/2021/formulas/physics/college/ofeth8jgtaxo9ks4rwkmehpqfbwuxlbgtr.png)
So
![w_c = 2 *0.00211](https://img.qammunity.org/2021/formulas/physics/college/j6wmsbo0yoe2jk605w2ce82g5kqmedykam.png)
![w_c = 0.00422 \ m](https://img.qammunity.org/2021/formulas/physics/college/bufil9qvkk46dy80lbi2h8sob3u8jmegcp.png)