Question

Two coherent narrow slits emitting the sound of wavelength λ in the same phase are placed parallel to each other at a small separation of $2\mathrm{\lambda }.$ The sound is detected by moving a detector on the screen ∑ at a distance $\mathrm{D}(>>\mathrm{\lambda })$ from the slit ${\mathrm{S}}_1$as shown in figure (below). Find the distance $\mathrm{x}$such that the intensity at P is equal to the intensity at O.

Answer 1

Step 1: Given data:
Two coherent narrow slits S₁ and S₂ are given to emit the sound of wavelength$=\mathrm{\lambda }$

The separation between the two slits is given as= $2\mathrm{\lambda }.$

The distance at which the sound is detected by moving a detector on the screen ∑ from the slit ${\mathrm{S}}_1$is given as$=\mathrm{D}$

Step 2: Calculating the maximum intensity at P:

S₁ and S₂ are given in to be the same phase, At O, there will be maximum intensity.

Hence the maximum intensity at P, can be calculated as:

From right-angled triangles,

$$\begin{gathered} {\mathrm{\Delta S}}_1\mathrm{PO}\mathrm{and}{\mathrm{\Delta S}}_2\mathrm{PO} \\ {\left({\mathrm{S}}_1\mathrm{P}\right)}^2–{\left({\mathrm{S}}_2\mathrm{P}\right)}^2 \\ =({\mathrm{D}}^2+{\mathrm{x}}^2)–({{(\mathrm{D}-2\mathrm{\lambda })}^2+{\mathrm{x}}^2)}^2=4\mathrm{\lambda D}+4{\mathrm{\lambda }}^2 \\ =4\mathrm{\lambda D} \end{gathered}$$

If λ is small, then λ² is negligible.

$$\begin{gathered} ({\mathrm{S}}_1\mathrm{P}+{\mathrm{S}}_2\mathrm{P})({\mathrm{S}}_1\mathrm{P}–{\mathrm{S}}_2\mathrm{P})=4\mathrm{\lambda D} \\ \Rightarrow ({\mathrm{S}}_1\mathrm{P}–{\mathrm{S}}_2\mathrm{P})=\frac{4\mathrm{\lambda D}}{\left[2\sqrt{({\mathrm{x}}^2+{\mathrm{D}}^2)}\right]}=\mathrm{n\lambda } \\ \Rightarrow \frac{2\mathrm{D}}{\left[\sqrt{({\mathrm{x}}^2+{\mathrm{D}}^2)}\right]}=\mathrm{n} \\ \Rightarrow {\mathrm{n}}^2({\mathrm{x}}^2+{\mathrm{D}}^2)=4{\mathrm{D}}^2 \\ \Rightarrow {\mathrm{n}}^2{\mathrm{x}}^2=4{\mathrm{D}}^2-{\mathrm{n}}^2{\mathrm{D}}^2 \\ \Rightarrow {\mathrm{n}}^2{\mathrm{x}}^2={\mathrm{D}}^2(4-{\mathrm{n}}^2) \\ \mathrm{or},\mathrm{x}=(\mathrm{D}/\mathrm{n})\sqrt{(4-{\mathrm{n}}^2)} \end{gathered}$$

For constructive interference, path difference is $n\lambda$.

Step 3: Calculating distance $\mathrm{x}$at which the intensity at P is equal to the intensity at O:

When $\mathrm{n}=1,\mathrm{x}=\sqrt{3\mathrm{D}}$

When $\mathrm{n}=2,\mathrm{x}=0$

Thus, when$\mathrm{x}=\sqrt{3\mathrm{D}}$, the intensity at P will be equal to the intensity at O.

Hence, the distance $\mathrm{x}$at which the intensity at P is equal to the intensity at O is given as $\sqrt{3\mathrm{D}}$.