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# Physics/Diffraction pattern

Question
Light of wavelength 630 nm from a distant source is falling on a slit of width 0.5 mm. The diffraction
pattern is observed on a screen 3.00 m away. What is the width of the central maximum?

The equation predicting the interference pattern produced by a single slit is:
n*lambda=d*sin(theta)
where theta is the angle at which the nth node appears. To find the width of the central maximum one will need to determine the distance to the first node and then double the distance to determine the total width of the central antinode.
Because the angle of the interference pattern in this case is so small the small angular approximation can be used to replace the sine of the angle with the angle theta in radians:
n*lambda=d*sin(theta)=d*theta=d*x/L
where x is the location on the screen of the first node and L is the distance to the screen.
In this case the given inputs are: n=1, lambda=630nm=630x10^(-9)m, d=0.5mm=5.0x10(-4)m, L=3.0m and x=the distance to the first node.
Substituting:
1*630x10^(-9)m=5.0x10(-4)m*x/3.0m
Solving for x:
x=0.00378m
Finally, double this number to determine the width W of the central maximum.
W=2*0.00378m=0.00756m=7.56mm

Physics