写一个复杂的计算衍射效率的python代码
时间: 2024-04-19 20:27:34 浏览: 16
下面是一个复杂的计算衍射效率的Python代码示例,该代码使用了一些数学函数和计算方法,供参考:
```python
import math
def sinc(x):
if x == 0:
return 1
else:
return math.sin(x) / x
def ksi(thetai, thetar, wavelength):
w = 12
wr = math.radians(w)
thetair = math.radians(thetai)
thetarr = math.radians(thetar)
result = (1/wavelength) * math.tan(wr) * (math.cos(thetair) + math.cos(thetarr))
return result
def calculate_diffraction_efficiency(a, wavelength, p, q, thetai):
thetai = math.radians(thetai)
sthai = math.sin(thetai)
cthetai = math.cos(thetai)
faii = 0
faiir = math.radians(faii)
sfaii = math.sin(faiir)
cfaii = math.cos(faiir)
for j in p:
for k in q:
print("衍射级次为:p:", j, ", q:", k)
x = solve([sin(thetarr)*cos(fairr) - sthai*cfaii - j/a*wavelength,
sin(thetarr)*sin(fairr) - sthai*sfaii - k/a*wavelength],
[thetarr, fairr])
for f in range(0, 2):
thetar = math.degrees(x[f][0])
fair = math.degrees(x[f][1])
if thetar < -90 or thetar > 90:
continue
if fair < -90 or fair > 90:
continue
n1 = a * (j/a - ksi(thetai, thetar, wavelength) / math.sqrt(2))
n11 = sinc(n1)
part1 = math.pow(n11, 2)
n2 = a * (k/a - ksi(thetai, thetar, wavelength) / math.sqrt(2))
n22 = sinc(n2)
part2 = math.pow(n22, 2)
n = part1 * part2
print("衍射效率为", n*100, "%")
# 示例使用
a = 13.68
wavelength = 10
p = [-1, 0]
q = [-1, 0]
thetai = 24
calculate_diffraction_efficiency(a, wavelength, p, q, thetai)
```
请注意,这只是一个示例代码,并不完整,需要根据您的具体需求进行调整和补充。其中,`calculate_diffraction_efficiency()`函数用于计算衍射效率,根据提供的参数进行计算,并输出结果。您可以根据自己的计算公式和逻辑进行修改。