Respuesta :
Answer:
The total number of wavelength is 35180.
Explanation:
Given that,
Thickness = 2.50 mm
Index of refraction =1.40
Wavelength = 540 nm
Distance = 1.80 cm
We need to calculate the wavelength in the medium
Using formula of wavelength
[tex]\lambda=\dfrac{\lambda_{air}}{n}[/tex]
[tex]\lambda=\dfrac{540\times10^{-9}}{1.40}[/tex]
[tex]\lambda=385.7\times10^{-9}\ m[/tex]
[tex]\lambda=385.7\ nm[/tex]
We need to calculate the number of wavelength
Using formula for number of wavelength
[tex]N_{1}=\dfrac{t}{\lambda}[/tex]
Put the value into the formula
[tex]N_{1}=\dfrac{2.50\times10^{-3}}{385.7\times10^{-9}}[/tex]
[tex]N_{1}=6.48\times10^{3}[/tex]
We need to calculate the distance traveled by light in air
Using formula for distance
[tex]D=d-t[/tex]
Put the value into the formula
[tex]D=1.80\times10^{-2}-2.50\times10^{-3}[/tex]
[tex]D=1.55\times10^{-2}\ m[/tex]
We need to calculate the number of wavelength
Using formula for number of wavelength
[tex]N_{2}=\dfrac{D}{\lambda_{air}}[/tex]
Put the value into the formula
[tex]N_{2}=\dfrac{1.55\times10^{-2}}{540\times10^{-9}}[/tex]
[tex]N_{2}= 2.87\times10^{4}[/tex]
We need to calculate the total number of wavelength
[tex]N=N_{1}+N_{2}[/tex]
Put the value into the formula
[tex]N=6.48\times10^{3}+2.87\times10^{4}[/tex]
[tex]N=35180[/tex]
Hence, The total number of wavelength is 35180.
In the given case, The numbers of wavelengths are there between the source and the screen is - 35185.2
The wavelength is the distance between corresponding points of two consecutive waves.
Given:
[tex]t = 2.50 \ mm = 2.50\times 10^{-3} \ m\\n_{glass} = 1.40\\\lambda_v = 540 \ nm\\ = 540\times 10^{-9} \ m\\d = 180 \ cm\\ = 1.80\times 10^{-2} \ m\\[/tex]
Solution:
wavelength =>
[tex]\Lambda = \frac{\lambda_v}{n_{glass}}\\ = \frac{540\times 10^{-9} \ m}{1.40}\\ = 385.71\times 10^{-9} \ m\\[/tex]
Number of waves =>
[tex]N_1 = \frac{t}{\lambda}\\ = \frac{2.50\times 10^{-3} \ m}{385.71\times 10^{-9} \ m}\\ = 6.481\times 10^3\\[/tex]
Distance :
[tex]D = d-t\\ = 1.80\times 10^{-2}-2.50\times 10^{-3}\\ = 1.55\times 10^{-2} \ m\\[/tex]
Thus, the
[tex]N_2 = \frac{D}{\lambda_v}\\ = \frac{1.55\times 10^{-2}}{540\times 10^{-9}}\\ = 2.87\times 10^4 \\[/tex]
The total number of wavelengths:
[tex]N = N_1+N_2\\ = 6.481\times 10^3+2.87\times 10^4\\ = 35185.2[/tex]
Thus, the numbers of wavelengths are there between the source and the screen is - 35185.2
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