Answer:
are you talking about links?
Step-by-step explanation:
yeah they can get annoying
Answer:
56.39 nm
Step-by-step explanation:
In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,
![2t=(m+\frac{1}{2} ).\frac{\lambda}{n}](https://tex.z-dn.net/?f=2t%3D%28m%2B%5Cfrac%7B1%7D%7B2%7D%20%29.%5Cfrac%7B%5Clambda%7D%7Bn%7D)
where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...
Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.
In order to have its longest constructive reflection at the red end (700 nm)
![t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm](https://tex.z-dn.net/?f=t_1%3D%28m%2B%5Cfrac%7B1%7D%7B2%7D%20%29.%5Cfrac%7B%5Clambda%7D%7B2.n%7D%5C%5C%20%5C%5C%20t_1%3D%5Cfrac%7B1%7D%7B2%7D%20.%5Cfrac%7B700%7D%7B%282%29%2A%281.33%29%7D%5C%5C%20%5C%5C%20t_1%3D131.58%5C%20nm)
Here we take m=0.
Similarly for the constructive reflection at the blue end (400 nm)
![t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm](https://tex.z-dn.net/?f=t_2%3D%28m%2B%5Cfrac%7B1%7D%7B2%7D%20%29.%5Cfrac%7B%5Clambda%7D%7B2.n%7D%5C%5C%20%5C%5C%20t_2%3D%5Cfrac%7B1%7D%7B2%7D%20.%5Cfrac%7B400%7D%7B%282%29%2A%281.33%29%7D%5C%5C%20%5C%5C%20t_2%3D75.19%5C%20nm)
Hence the thickness difference should be
![t_1-t_2=131.58-75.19=56.39 \ nm](https://tex.z-dn.net/?f=t_1-t_2%3D131.58-75.19%3D56.39%20%5C%20nm)
Answer:
thanks im not smart
Step-by-step explanation:
sn i knew that...
Step-by-step explanation:
x+0.6+0.3-0.2=14.2
x+0.7=14.2
x=14.2-0.7
x=13.5