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A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic light of 630-nm wavelength is used. The same slit, illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum. What is the wavelength of this new light?

Respuesta :

Answer:

[tex]\lambda = 424.4 nm[/tex]

Explanation:

As we know by the formula of diffraction

[tex]a sin\theta = N\lambda[/tex]

so we have

a = slit size

[tex]\theta [/tex] = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

[tex]a sin40 = 1(630 \times 10^{-9})[/tex]

[tex]a = 9.8 \times 10^{-7} m[/tex]

Now for another wavelength second minimum is at 60 degree angle

[tex]a sin60 = 2 \lambda[/tex]

[tex](9.8 \times 10^{-7}) sin60 = 2 \lambda[/tex]

[tex]\lambda = 424.4 nm[/tex]

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