Respuesta :
To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.
By definition in the principle of superposition, light interference is defined as
[tex]Y = \frac{m\lambda R}{d}[/tex]
Where,
d = Separation of the two slits
[tex]\lambda = Wavelength[/tex]
R = Distance from slit to screen
m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.
Y = Distance from central spot to fringe.
Re-arrange the equation to find \lambda we have that
[tex]\lambda = \frac{Yd}{mR}[/tex]
Our values are gives according the problem as,
[tex]Y = 10*10^{-2}m[/tex]
[tex]d = 3*10^{-5}[/tex]
m = 7 (The seventh bright fringe)
R = 2m
[tex]\lambda = \frac{(10*10^{-2})(3*10^{-5})}{7*2}[/tex]
[tex]\lambda = 214nm[/tex]
Therefore the wavelength of the light passing through the slits is 214nm
The wavelength of the light passing through the slit is 214 nm.
What is the wavelength?
The wavelength is the distance between identical points in the adjacent cycles of a waveform.
Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.
The wavelength of the light passing through the slit is calculated as given below.
[tex]\lambda = \dfrac {sd}{mr}[/tex]
[tex]\lambda = \dfrac {10\times 10^{-2} \times 3.00 \times 10^{-5}}{7 \times 2}[/tex]
[tex]\lambda = 2.14 \times 10^{-7}\;\rm m[/tex]
[tex]\lambda = 214 \;\rm nm[/tex]
Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.
To know more about the wavelength, follow the link given below.
https://brainly.com/question/7143261.
