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July 31, 2014

July 31, 2014

Posted by **Jim** on Saturday, March 31, 2012 at 4:11pm.

- Physics -
**drwls**, Saturday, March 31, 2012 at 5:37pmThe condition for an intensity minimum is b sin á = k ë

where b is the slit width.

For 474 nm, there are minima at

sin á = 6.21*10^-3 (k=1)

1.242*10^-3 (k=2)

1.864*10^-2 (k=3)

2.484*10^-2 (k=4)

For 632 nm, there are minima at:

sin á = 8.283*10^-3 (k=1)

1.657*10^-3 (k=2)

2.485*10^-2 (k=3)

Clearly, the k=4 minimum of ë = 474 nm radiation coincides with the k = 3 minimum of 632 nm radiation. The distance from the central maximum is

d = 1.30 m*sin á = 3.23*10^-2 m

= 3.23 cm

- Physics -
**drwls**, Saturday, March 31, 2012 at 5:55pmFor further reading, try

http://www.walter-fendt.de/ph14e/singleslit.htm

The a' symbol in my answer above was meant to be the diffraction angle alpha, and the e with two dots was supposed to be the wavelength, lambda

- Physics -
**Elena**, Saturday, March 31, 2012 at 6:06pmLet’s use the condition of diffraction minimum for one split of the width b

b•sinα =k1•λ1

b•sinα =k2•λ2,

Since we have superposition of two maxima b•sinα is the same for two wavelengths,

k1•λ1= k2•λ2,

λ1/λ2 = k2/k1,

632/474 = 4/3.

Therefore, k1=3, k2=4.

Now sinα = k1•λ1/b =3•632•10^-9/ 7.63•10^-5 =0.0248.

As the angle is very small tanα = sinα,

x=L• tanα =L• sinα =1.3•0.0248 =0.0323 m = 3.23 cm

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