Answer:
276.19nm
Explanation:
To find the other wavelength you use the following condition for the diffraction of both wavelengths:
[tex]m_1\lambda_1=asin\theta_1\\\\m_2\lambda_2=asin\theta_2\\\\[/tex] ( 1 )
λ1=410nm
m=1 for wavelength 1
m=2 for wavelength 2
a: width of the slit
θ1: angle of the first minimum
θ2: angle of the second minimum
you divide both equations and you obtain:
[tex]\frac{m_1\lambda_1}{m_2\lambda_2}=\frac{sin\theta_1}{sin\theta_2}\\\\\lambda_2=\frac{sin\theta_2}{sin\theta_1}\frac{m_1\lambda_1}{m_2}\\\\\lambda_2=\frac{sin60\°}{sin40\°} \frac{(1)(410nm)}{2}=276.19nm[/tex]
hence, the wavelength of the second monochromatic wave is 276.19nm
Answer:
The wavelength of the second monochromatic light is [tex]\lambda = 277 nm[/tex]
Explanation:
From the question we are told that
The angle of the first minimum is [tex]\theta_1 = 40^o[/tex]
The wavelength of the first monochromatic light is [tex]\lambda_1 = 410 \ nm[/tex]
The angle of the second minima is [tex]\theta_2 = 60^o[/tex]
For the first minima the distance of separation of diffraction patterns is mathematically represented as
[tex]a = \frac{\lambda_1 }{sin \theta_1}[/tex]
Substituting values
[tex]a = \frac{410 *10^{-9}}{sin (40) }[/tex]
[tex]a = 638 nm[/tex]
The distance between two successive diffraction is constant for the same slit
Thus the wavelength of the second light is
[tex]\lambda = \frac{a * sin (60)}{2}[/tex]
Substituting value
[tex]\lambda = \frac{638 * sin (60)}{2}[/tex]
[tex]\lambda = 277 nm[/tex]
