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Diffraction Grating Formula

A diffraction grating is made by making many parallel scratches on the surface of a flat piece of some transparent material. It is possible to put some large number of scratches per cm on the material. For example, the grating to be used has 6,000 lines per cm on it. The scratches are opaque but the areas between the scratches can transmit the light through. Thus, a diffraction grating becomes a multitude for the source with parallel slit, when light falls upon it. In this topic, a student will learn the diffraction grating formula with examples. Let us learn the concept!

Diffraction Grating Formula

Diffraction Grating Formula

Concept of the diffraction grating

A parallel bundle of the rays will fall on the grating. Rays and wavefront form an orthogonal set so the wavefront will be perpendicular to the rays and parallel to the grating. Here Huygens’ Principle is applicable.

According to it every point on a wavefront acts as a new source, and each transparent slit becomes a new source so cylindrical wavefront spread out from each.

If a peak falls on a valley consistently, then the waves cancel and no light exists at that point. Also, if peaks fall on peaks and valleys fall on valleys consistently, then the light is made brighter at that point. Diffraction is an alternative way to observe spectra other than a prism.

The formula for diffraction grating:

Consider two rays that emerge making the angle \(\theta\) with the straight through the line. Constructive interference will occur if the difference in their two path lengths is an integral multiple of their wavelength \(\lambda\) i.e.,

Now, n \(\lambda = d sin (\theta )\)

where n = 1, 2, 3, …


In this formula, \(\theta\) is the angle of emergence at which a wavelength will be bright. Also, d is the distance between slits. Obviously,

d = \(\frac {1} { N }\), where N is the grating constant, and it is the number of lines per unit length. Also, n is the order of grating, which is a positive integer, representing the repetition of the spectrum.

Furthermore, a complete spectrum could be observed for n = 1 and another complete spectrum for n = 2, etc., but at the larger angles.

Solved Examples

Q.1: A diffraction grating diffracts light at \(20 ^{\circ}\). What will be the wavelength of the light if the grating has 1000 lines per mm length?


Given parameters are

\(\theta = 20 ^{\circ}\).

N =1000 limes per mm length.

So, d =\(\frac {1}{N}\)

i.e. d = \(\frac {1}{1000}\) mm

i.e. d = \(10^{-6}\) m.

for n =1,

using formula , n \(\lambda = d sin (\theta )\)

\(\lambda = d sin (\theta\) putting n =1

= \(10^{-6} \times sin 20 ^{\circ}\)

= \(3.4 \times 10^{-5} m\)

Thus wavelength will be \(3.4 \times 10^{-5} m\)

Q.2: A diffraction grating has 36000 lines per cm and it separates a dark line at an angle of \(32 ^{\circ}\). Find the wavelength of the light.


Given values are,

Separation between the slits, d = \(\frac {1} {36000} = 2.77 \times 10^{−5} cm = 2.77 \times 10^{−7} m\)

\(\theta = 32 ^{\circ}\).

Wavelength will be:

n \(\lambda = d sin (\theta )\)

\(\lambda  = 2.77 \times 10^{−7} \times sin  32 ^{\circ}.            ( n = 1)\)

\(\lambda\) =  146.7 nm

Therefore, the wavelength of light will ne 146.7 nano meter.

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