Question 56: Waves Optics

Question

Two coherent monochromatic point sources $${S_1}$$ and $${S_2}$$ of wavelength $$\lambda  = 600\,nm$$ are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance $$d=1.8$$ $$mm.$$ This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $$\Delta 	heta .$$ Which of the following options is/are correct?

Accepted Answer

At any point P on circumference, the path difference, $$\Delta$$x = d sin$$	heta$$
At P
1
, $$	heta$$ = 0$$^\circ$$ $$\Rightarrow$$ $$\Delta$$x = 0
At P
2
, $$	heta$$ = 90$$^\circ$$ $$\Rightarrow$$ $$\Delta$$x = d
For constructive interference, $$\Delta$$x = n$$\lambda$$
$$	herefore$$ $$n = {d \over \lambda } = {{1.8\,mm} \over {600\,nm}} = {{1.8 	imes {{10}^{ - 3}}m} \over {600 	imes {{10}^{ - 9}}m}} = 3000$$
Since, n is integer, hence P
2
corresponds to bright spot and also it corresponds to maximum order of fringe.
Now, $$\Delta$$x = d sin$$	heta$$
d$$\Delta$$x = d cos$$	heta$$ d$$	heta$$ or R$$\lambda$$ = d cos$$	heta$$ d$$	heta$$
$$	herefore$$ $$d	heta  \propto {1 \over {\cos 	heta }}$$
Hence, as $$	heta$$ increases, cos $$	heta$$ decreases and consequently d$$	heta$$ increases.

Answer

The angular separation between two consecutive
bright spots decreases as we move from P1 to P2 along
the first quadrant.

Answer

At P2 the order of the fringe will be maximum.

Answer

A dark spot will be formed at the point P2.

Answer

The total number of fringes produced between P1 and
P2 in the first quadrant is close to 3000.

Waves Optics

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