The problem involves a laser with a wavelength of 632.8 nm passing through a single slit of width 0.30 mm, and the goal is to determine the width of the central maximum on a screen located 1.0 m behind the slit.
The discussion is ongoing, with participants providing guidance on how to proceed after calculating sin(theta). There is an exploration of trigonometric relationships to find the height corresponding to the central maximum, but no consensus has been reached on the next steps.
Participants express uncertainty about the process of converting sin(theta) values into angles and how to apply trigonometric properties to find the desired dimensions on the screen.
xswtxoj said:Homework Statement
A laser 632.8nm is sent through a single slit of widith 0.30 mm. Wats the width of the central maximum on a screen 1.0m in the back of the slit?
Homework Equations
asin(theta) = m lamda
The Attempt at a Solution
lamda: 632.8nm
0.30 mm= 3.0 x 10^-4 m
sin(theta)= m lamda/ a
sin(theta)= 1 (632.8x10^-9)/3.0 x 10^-4 m
= 2.10x10^-3 m
xswtxoj said:do i inverse sin 2.10 E -3? I'm nto quick sure what to do?