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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

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.