How Wide Is the Slit in a Diffraction Experiment with 450nm Light?

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The discussion centers on calculating the slit width in a diffraction experiment using 450nm light, where the central diffraction peak is 6cm wide at a distance of 2m. The relevant formulas include the relationship between the width of the slit, wavelength, and the angle of diffraction. An initial attempt at the solution incorrectly applied the formula for sine of the angle, leading to an erroneous calculation of the slit width. A correction was provided, indicating that the correct formula should be sin(theta) = m * lambda / w. The accurate calculation is essential for determining the slit width in this diffraction scenario.
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Homework Statement


When 450nm light falls on a slit, the central diffraction peak (central maxima) on a screen 2.0m away is 6cm wide. Calculate the slit width?
wavelength= 450 x 10^9m
L= 2m
y= 0.06m
w=?

Homework Equations


y/L = tantheta = sintheta
sintheta = (m+1/2)lambda / w

The Attempt at a Solution


I attempted at this, and got this answer... can anyone check to see if I did it right?
y/L = sintheta
0.06/2 = 0.03

sintheta = (m+1/2)lambda / w
0.03 = (1+1/2) (450 x 10^-9) / w
w= 2.25 x 10^-5m
 
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cassey said:
sintheta = (m+1/2)lambda / w
You are using a wrong formula. It should be ##\sin\theta = m\lambda / w##.
 
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