Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.63 x 10-5 m and strike a screen 1.30 m from the slit. Two diffraction patterns are formed on the screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern?

The 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

For 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

Let’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

To find the distance between the center of the diffraction patterns and the first occurrence of the spot where dark fringes overlap, we first need to determine the positions of the dark fringes for each wavelength separately.

The position of the dark fringes in a single-slit diffraction pattern can be determined using the formula:

y = (m * λ * L) / w

where:
y is the distance from the center of the diffraction pattern to the position of the dark fringe
m is the order of the fringe (0, 1, -1, 2, -2, etc.)
λ is the wavelength of the light
L is the distance between the slit and the screen
w is the width of the slit

For the first wavelength (632 nm), we can calculate the position of the dark fringe:

y1 = (m * λ1 * L) / w

Substituting the values:
λ1 = 632 nm = 6.32 x 10^-5 m
L = 1.30 m
w = 7.63 x 10^-5 m

For the second wavelength (474 nm), we can calculate the position of the dark fringe:

y2 = (m * λ2 * L) / w

Substituting the values:
λ2 = 474 nm = 4.74 x 10^-5 m

Now, we can find the distance between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern. This distance can be calculated by subtracting the position of the dark fringe for one wavelength from the position of the dark fringe for the other wavelength:

distance = |y1 - y2|

Now, let's plug in the values and calculate the distance:

distance = |((m * 6.32 x 10^-5 * 1.30) / (7.63 x 10^-5)) - ((m * 4.74 x 10^-5 * 1.30) / (7.63 x 10^-5))|

Since 'm' can be any integer value, we need to consider the absolute value to find the minimum distance. We can calculate this for various values of 'm' and take the minimum value.

The result will be the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern.