Finding Central Maximum Width for Laser

AI Thread Summary
To find the width of the central maximum for a laser passing through a single slit, the wavelength of the laser (632.8 nm) and the slit width (0.30 mm) are used in the formula sin(theta) = m * lambda / a. The calculated sin(theta) value is approximately 2.10 x 10^-3. To determine the angle theta, the inverse sine function should be applied to this value. Once theta is found, basic trigonometry can be used to calculate the height of the central maximum on the screen, noting that this height represents half the total width of the central maximum. Understanding these steps is crucial for solving the problem accurately.
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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
 
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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?
 

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