Question

Calculate the wavelength (in nm) of light that produces its first minimum at an angle of 21.0° when falling on a single slit of width 1.46 µm.

Answer 1

To solve this problem it is necessary to apply the related concepts to the principle of overlap, specifically to single slit diffraction experiment concept.

Mathematically this can be expressed as:

`dsin\theta = m\lambda`

Where,

d = Width of the slit

`\lambda =`Wavelength

`\theta =` Angle relative to the original direction of the light

m = Any integer which represent the order of the equation (number of repetition of the spectrum)

To solve the problem we need to rearrange the equation and find the wavelength

`\lambda = (dsin\theta)/(m)`

Our values are given as,

`d = 1.46\mu m = 1.46*10^(-6)m`

`\theta = 21\°`

`m = 1`

Replacing in our equation we have,

`\lambda = (dsin\theta)/(m)`

`\lambda = ((1.46*10^(-6))sin(21))/(1)`

`\lambda = 5.232*10^(-7)m`

`\lambda = 523.2nm`

Therefore the wavelength is 523.2nm