Question

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?

Answer 1

`\lambda = 424.4 nm`

Step-by-step explanation:

As we know by the formula of diffraction

`a sin\theta = N\lambda`

so we have

a = slit size

`\theta` = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

`a sin40 = 1(630 * 10^(-9))`

`a = 9.8 * 10^(-7) m`

Now for another wavelength second minimum is at 60 degree angle

`a sin60 = 2 \lambda`

`(9.8 * 10^(-7)) sin60 = 2 \lambda`

`\lambda = 424.4 nm`