Fringe Shift

Fringe shift due to thin film

If a transparent sheet of refractive index (μ) and thickness (t) is introduced in the path of one of the interfering light waves then, the optical path difference between the light waves will become

$\large \Delta x’ = \Delta x \pm (\mu – 1)t $

$\large \Delta x’ = \frac{y d}{D} \pm (\mu – 1)t $

where slits are separated by d.

In absence of the sheet, the position of central maxima is given by,

y = 0

The new position of the central maxima will be for Δx’ = 0

$\large \ \frac{y d}{D} = (\mu – 1)t $ ; (Taking -ve sign)

$\large y = \frac{(\mu – 1)t D}{d} $

$\large y = \frac{\beta}{\lambda}(\mu-1)t$

Therefore the entire fringe pattern will shift towards the side where the film is placed and there will be no change in the fringe width.

Illustration : Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10-5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet?

Solution : As derived earlier,

Total fringe shift $\large = \frac{\beta}{\lambda}(\mu-1)t$

As fringe width = β

Number of fringe that will shift $\large = \frac{Total \; fringe \; shift}{fringe \; width}$

$\large = \frac{\beta}{\lambda}(\mu-1)t / \beta$

$\large = \frac{(\mu-1)t}{\lambda}$

$\large = \frac{(1.6-1)1.8 \times 10^{-5}}{600 \times 10^{-9}}$

= 18

Illustration : A thin sheet of glass (μ= 1.520) is introduced normally in the path of one of the two interfering waves. The central bright fringe is observed to shift to the position originally occupied by the fifth bright fringe. If  λ = 5890A°, find the thickness of the glass sheet.

Solution : x0 = displacement of images = 5 β

$\large \frac{\beta}{\lambda}(\mu-1)t = 5 \beta$

$\large t = \frac{5 \lambda}{(\mu – 1)}$

$\large t = \frac{5 \times 5890 \times 10^{-10}}{(1.520 – 1)}$

t = 5.66 × 10-4 cm

Exercise : In a double slit experiment, Io is the intensity of the central bright fringe obtained with monochromatic light of λ = 6000 A°. Determine the intensity at a distance of 4.8 × 10-5m from the central maximum if the separation between the slits is 0.25 cm and the distance between the screen and the double slit is 1.20 m.

Exercise : Interference fringes are produced by a double slit arrangement and a piece of plane parallel glass of refractive index 1.5 is interposed in one of the interfering beam. If the fringes are displaced through 30 fringe widths for light (λ = 6 × 10-5 cm ), find the thickness of slab.

Also Read :

∗ Phase difference & Path difference
∗ Interference , Constructive & Destructive interference
∗ Young’s double slit experiment
∗ Calculation of fringe width in YDSE
∗ Angular fringe width
∗ Change in fringe width due to change in medium
∗Diffraction of Light

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