Question
Two sources $${S_1}$$ and $${S_2}$$ emitting coherent light waves of wavelength $$\lambda $$ in the same phase are situated as shown. The distance $$OP,$$ so that the light intensity detected at $$P$$ is equal to that at $$O$$ is
Two sources $${S_1}$$ and $${S_2}$$ emitting coherent light waves of wavelength $$\lambda $$ in the same phase are situated as shown. The distance $$OP,$$ so that the light intensity detected at $$P$$ is equal to that at $$O$$ is
A.
$$D\sqrt 2 $$
B.
$$\frac{D}{2}$$
C.
$$D\sqrt 3 $$
D.
$$\frac{D}{{\sqrt 3 }}$$
Answer :
$$D\sqrt 3 $$
Solution :
Referring to the figure, the path difference between the two waves starting from $${S_1}$$ and $${S_2}$$ tums out to be $$\left( {2\lambda \cos \theta } \right) = n\lambda $$ where $$n$$ is taken as $$1$$ to get the point of maximum intensity which is the same as a point $$O.$$
Therefore, the above relation gives $$\cos \theta = \frac{1}{2}$$ so that $$\theta = {60^ \circ }$$ and $$\tan \theta = \frac{{PO}}{D} = \sqrt 3 ,$$ giving $$PO = D\sqrt 3 .$$
Referring to the figure, the path difference between the two waves starting from $${S_1}$$ and $${S_2}$$ tums out to be $$\left( {2\lambda \cos \theta } \right) = n\lambda $$ where $$n$$ is taken as $$1$$ to get the point of maximum intensity which is the same as a point $$O.$$
Therefore, the above relation gives $$\cos \theta = \frac{1}{2}$$ so that $$\theta = {60^ \circ }$$ and $$\tan \theta = \frac{{PO}}{D} = \sqrt 3 ,$$ giving $$PO = D\sqrt 3 .$$