Question

Figure illustrates the interference experiment with Fresnel mirrors. The angle between the mirrors is $α = 12^{'},$ the distances from the mirrors' intersection line to the narrow slit $S$ and the screen $S c$ are equal to $r = 10.0 cm$ and $b = 130 cm$ , respectively. The wavelength of light is $λ = 0.55 μm$ . Find:
(a) the width of a fringe on the screen and the number of possible maxima;
(b) the shift of the interference pattern on the screen, when the slit is displaced by $δ l = 1.0 mm$ along the arc of radius $r$ , with centre at the point $O$ ;
(c) at what maximum width $δ_{\max}$ of the slit, the interference fringes on the screen are still observed sufficiently sharp.

Accepted Answer

(a) In the given figure, $S'', S'$ and $S$ are lying on a circle of radius $r$ .

Here, $S'$ and $S''$ behave as coherent sources. Since $α$ is very small, so line $S' S''$ is parallel to the screen. So, the formula for fringe width $β = \frac{D λ}{d}$ is applicable.

Clearly from the figure, $\tan α = \frac{d}{2 r}$

But $α$ is very small,

$∴ \tan α = α = \frac{d}{2 r}$

$∴ d = 2 r α$

$∴$ The fringe width is

$β = \frac{D λ}{d} = \frac{(b + r) λ}{2 r α} . . . (1)$

Note: Students are advised to remember this result.

Given: $α = 12^{'},$ $α = 12 minute = {(\frac{12}{60})}^{\circ} = \frac{12}{60} \times \frac{π}{180} rad$

Putting the values, we get fringe width

$β = 1.1 mm$

Now, the arrangement is similar to YDSE. So, the number of fringes must be odd.

The number of fringes is given by $n = 2 [n_{1}] + 1$

where, $n =$ Total number of fringes

$n_{1} =$ Number of fringes on the either side of the central bright fringe.

From the figure, the fringe pattern on the screen is formed from point $A$ to point $B$ , and $C$ is the central position of the screen.

Let $C A = C B = l$

$∴ \tan α = α = \frac{l}{b}$

$\Rightarrow l = b α$

$∴$ Number of fringes on either side of point $C$ is $n_{1} = \frac{l}{β}$

$∴ n_{1} = \frac{b α}{β} \Rightarrow n_{1} = \frac{130 \times 10^{- 2}}{1.1 \times 10^{- 3}} \times \frac{12}{60} \times \frac{π}{180} \Rightarrow n_{1} = 4.15$

Hence, the maximum number of fringes is

$n = 2 [n_{1}] + 1 = 2 [4.15] + 1 = 2 \times 4 + 1$

$\Rightarrow n = 9$

(b) When an incident ray is rotated through an angle $δ θ,$ then image or reflected ray is rotated with the same angle in the opposite sense.

When slit is shifted, then incident ray on the mirror is rotated through $δ θ = \frac{δ l}{r}$ , in clockwise direction. So, reflected rays are rotated through angle $δ θ$ , in anticlockwise direction. Due to this, $S'$ shifts to $S'_{1}, S''$ shifts to $S''_{1} .$ $A$ shifts to $A'$ and $B$ shifts to $B'$ , as shown in the figure.

Hence, shifting in fringe pattern is

$A A' = B B' = b δ θ = \frac{b δ l}{r} . . . (2)$

Putting the values, we get

$A A' = B B' = 13 mm .$

(c) We can think of slit width $δ$ as two narrow slits of YDSE separated by a distance $δ$ .

The fringe pattern observed on the screen will not be sharp if the pattern due to the two narrow slits are $\frac{β}{2}$ (half of the fringe width) apart, because then the maxima due to one of the slits will overlap with the minima of the other slit. Thus, we have,

$\frac{b δ_{\max}}{r} = \frac{β}{2}$ [using equation $(2)$ ]

$\Rightarrow \frac{b δ_{\max}}{r} = \frac{(b + r) λ}{4 r α}$ [using equation $(1)$ ]

$\Rightarrow δ_{\max} = (1 + \frac{r}{b}) \frac{λ}{4 α}$

Putting the values, we get,

$δ_{\max} = 43 \times 10^{- 6} m = 43 μm$

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