Question

In a double-slit interference experiment, the slit separation is
2.29 μm, the light wavelength is 532 nm, and the separation between
the slits and the screen is 4.42 m. **(a)** What is
the angle between the center and the third side bright fringe? If
we decrease the light frequency to 94.8% of its initial value,
**(b)** does the third side bright fringe move along
the screen toward or away from the pattern's center and
**(c)** how far does it move?

Answer #1

**given**

**d = 2.29 micro m = 2.29*10^-6 m
lamda = 532 nm = 532*10^-9 m
R = 4.42 m**

**a) for third bright fringe(m=3), d*sin(theta) =
m*lamda**

**sin(theta) = m*lamda/d**

**sin(theta) = 3*532*10^-9/(2.29*10^-6)**

**theta = sin^-1(0.6969)**

**= 44.2 degrees**

**b) if the frequency decreases wavelength
increases.**

**so, the third bright fringe moves away from the
pattern's center.**

**c) actual place of third bright fringe from the center, y3
= m*lamda*R/d**

**= 3*532*10^-9*4.42/(2.29*10^-6)**

**= 3.080 m**

**when frequency decreases to 94.8%,**

**wavelegth, lamda' = lamda/0.948**

**y3' = m*lamda'*R/d**

**= 3*(532*10^-9/0.948)*4.42/(2.29*10^-6)**

**= 3.249 m**

**so, y3' - y3 = 3.249 - 3.080**

**= 0.169 m
<<<<<<<<<<<<---------------Answer**

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