First slide

Diffraction

Question

At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

Difficult

A

$π radian$
B

$\frac{π}{8} radian$
C

$\frac{π}{4} radian$
D

$\frac{π}{2} radian$

Solution

The situation is shown in the figure.

In figure A and B represent the edges of the slit AB of width a and C represents the midpoint of the slit. For the first minimum at P,
For the first minimum at P,
$a \sin θ = λ . . . . (i)$
$where λ is the wavelength of light.$
$The path difference between the wavelets from A to C is$
$Δ x = \frac{a}{2} \sin θ = \frac{1}{2} (a \sin θ)$
$= \frac{λ}{2} (using (i))$
$The corresponding phase difference Δ ϕ is$
$Δ ϕ = \frac{2 π}{λ} Δ x = \frac{2 π}{λ} \times \frac{λ}{2} = π$

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