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Thermodynamics

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Oblique Incidence

Let a parallel beam of light be incident obliquely on the grating surface at an angle of incidence i.

The path difference between the secondary waves passing the point A and C = FC + CE.

Here, AB = a, the width of the slit and BC = b, the width of the opaque portion.

Then, from the Δ AFC

FC = (a + b) sin i and from Δ ACE

CE = (a + b) sin θ

∴ FC + CE = (a + b) (sin θ + sin i)           (i)

Equation (i) holds good if the beam is diffracted upwards. Fig. illustrates the diffraction of the beam downwards. In this case the path difference

= (a + b) [sin θ – sin i]                           (ii)

For the nth primary maximum

(a + b) [sin θn + sin i] = nλ                    (iii)

The deviation of the diffraction beam = θn + 1

For the deviation (θn + i) to be minimum, must be minimum. This is possible if the value of is maximum i.e.

Thus, the deviation produced in the diffracted beam is minimum when the angle of incidence is equal to the angle of diffraction. Let Dm be the angle of minimum deviation.

Then, Dm = θn + i

But, θn = i

Equation (v) refers to the principal maximum of the nth order for a wavelength λ.