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

$\lambda = \frac{dsin\theta}{m}$

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,

$\lambda = \frac{dsin\theta}{m}$

$\lambda = \frac{(1.46*10^{-6})sin(21)}{1}$

$\lambda = 5.232*10^{-7}m$

$\lambda = 523.2nm$

Therefore the wavelength is 523.2nm