Question 51: Wave Optics

Question

In a Young double slit experiment, the wavelength of incident light is $6000 \mathop {\rm{A}}\limits^{\rm{o}}$, the separation between slits $S_1$ and $S_2$ is 5 cm and the distance between slits plane and screen is 50 cm , as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is $\_\_\_\_$ Å.

Accepted Answer

When two waves of intensities $I_1$ and $I_2$ interfere with a phase difference $\phi$, the resultant intensity $I_R$ is given by:
$$ \mathrm{I}_{\mathrm{R}}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi $$
Since the slits are identical, the intensities due to individual slits are equal ( $\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}_0$ ). The formula simplifies to:
$$ \mathrm{I}_{\mathrm{R}}=\mathrm{I}_0+\mathrm{I}_0+2 \sqrt{\mathrm{I}_0^2} \cos \phi $$
$$\Rightarrow $$ $\mathrm{I}_{\mathrm{R}}=2 \mathrm{I}_0(1+\cos \phi)$
Using the trigonometric identity $1+\cos \phi=2 \cos ^2\left(\frac{\phi}{2}\right)$
$$ \mathrm{I}_{\mathrm{R}}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right) $$
It is given that the resultant intensity at point P is equal to the intensity of an individual slit:
$$ \mathrm{I}_{\mathrm{R}}=\mathrm{I}_0 $$
Substituting this into equation:
$$ \mathrm{I}_0=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}\right) $$
$$\Rightarrow $$ $\cos ^2\left(\frac{\phi}{2}\right)=\frac{1}{4}$
$$\Rightarrow $$ $$ \cos \left(\frac{\phi}{2}\right)= \pm \frac{1}{2} $$
For $\cos \left(\frac{\phi}{2}\right)=\frac{1}{2} \Rightarrow \frac{\phi}{2}=\frac{\pi}{3} \Rightarrow \phi=\frac{2 \pi}{3}$
For $\cos \left(\frac{\phi}{2}\right)=-\frac{1}{2} \Rightarrow \frac{\phi}{2}=\pi-\frac{\pi}{3} \Rightarrow \frac{\phi}{2}=\frac{2 \pi}{3} \Rightarrow \phi=\frac{4 \pi}{3}$
As we take phase difference in the range $0 \leq \phi \leq \pi$, therefore the required phase difference between interfering waves will be $\phi=\frac{2 \pi}{3}$.
For other values of $\phi$ which satisfy the condition $\cos ^2\left(\frac{\phi}{2}\right)=\frac{1}{4}$, the path difference will be same. So, we take the principle solution in the range $0 \leq \phi \leq \pi$.
The phase difference is related to the path difference by the formula:
$$ \phi=\frac{2 \pi}{\lambda}(\Delta x) $$
$$\Rightarrow $$ $$ \Delta \mathrm{x}=\frac{\phi \cdot \lambda}{2 \pi} $$
The wavelength of light is given as $\lambda=6000 ÃÂ$ and $\phi=\frac{2 \pi}{3}$ :
$$ \Delta \mathrm{x}=\frac{\left(\frac{2 \pi}{3}\right) \cdot 6000 ÃÂ}{2 \pi} $$
$$\Rightarrow $$ $\Delta \mathrm{x}=\frac{6000}{3} ÃÂ$
$$\Rightarrow $$ $$ \Delta \mathrm{x}=2000 ÃÂ $$
Therefore, the path difference between interfering waves is $2000 ÃÂ$.
Thus, the correct option is (C).

Answer

4000

Answer

3000

Answer

2000

Answer

1000

Wave Optics

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