# Finding Central Maximum Width for Laser

• xswtxoj
In summary, the problem involves sending a 632.8nm laser through a single slit with a width of 0.30 mm. Using the equation asin(theta) = m lamda, it is determined that the width of the central maximum on a screen 1.0m behind the slit is 2.10x10^-3 m. In order to find the unknown height of the screen, the inverse sin function is used to find the angle theta.

## 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:

## 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

Ok you're on the right track, you found sintheta. Now use your inverse sin function and find the angle theta. Once you have that, you know the distance to the screen. Can you find a basic trig property to help you find the unknown "height" if you will of the screen? (remember, this height you find will only be half the width of the central maximum)

do i inverse sin 2.10 E -3? I'm nto quick sure what to do?

xswtxoj said:
do i inverse sin 2.10 E -3? I'm nto quick sure what to do?

You don't know how to turn SinTheta = x into an angle?

What Theta corresponds to SinTheta = 1/sqrt2 for example?