Question 443: Wave 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 P
1
to P
2
along
the first quadrant.

Answer

At P
2
the order of the fringe will be maximum.

Answer

A dark spot will be formed at the point P
2
.

Answer

The total number of fringes produced between P
1
and
P
2
in the first quadrant is close to 3000.

Wave Optics

Choose the correct answer: