Find the ratio of intensities of two points P and Q on a screen in a Young's double slit experiment when waves from sources S1 and S2 have phase difference of (i)lambda/6 and (ii)lambda/12 respectively.

In Young's double-slit experiment, the intensity at any point on the screen is determined by the amplitude of the resultant wave at that point. The amplitudes depend on the phase difference \(\Delta \phi\) between the waves reaching the point.

The intensity \(I\) is proportional to the square of the resultant amplitude. The resultant amplitude depends on the individual amplitudes \(A_1\) and \(A_2\) of the two waves, and their phase difference \(\Delta \phi\). Let us derive the ratio of intensities for the given phase differences.

### General Formula
The resultant amplitude \(A_R\) is given by:
\[
A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\Delta \phi)}
\]
If the amplitudes of the two waves are equal (\(A_1 = A_2 = A\)), the resultant amplitude becomes:
\[
A_R = A\sqrt{2(1 + \cos(\Delta \phi))} = A\sqrt{4\cos^2\left(\frac{\Delta \phi}{2}\right)}
\]
\[
A_R = 2A\cos\left(\frac{\Delta \phi}{2}\right)
\]

The intensity \(I\) is proportional to the square of the resultant amplitude:
\[
I \propto A_R^2 = 4A^2\cos^2\left(\frac{\Delta \phi}{2}\right)
\]

### (i) Phase Difference \(\Delta \phi = \frac{\lambda}{6}\)
The phase difference in terms of wavelength is related to the actual phase difference \(\Delta \phi\) using:
\[
\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}
\]
The intensity at point \(P\) is:
\[
I_P \propto 4A^2\cos^2\left(\frac{\pi}{6}\right)
\]
Using \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\):
\[
I_P \propto 4A^2\left(\frac{\sqrt{3}}{2}\right)^2 = 4A^2 \cdot \frac{3}{4} = 3A^2
\]

### (ii) Phase Difference \(\Delta \phi = \frac{\lambda}{12}\)
The phase difference is:
\[
\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{12} = \frac{\pi}{6}
\]
The intensity at point \(Q\) is:
\[
I_Q \propto 4A^2\cos^2\left(\frac{\pi}{12}\right)
\]

Using \(\cos\left(\frac{\pi}{12}\right)\) and simplifying numerically or symbolically:
\[
\cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}}
\]
\[
I_Q \propto 4A^2 \cdot \left(\frac{\sqrt{2 + \sqrt{3}}}{2}\right)^2 = 4A^2 \cdot \frac{2 + \sqrt{3}}{4} = A^2(2 + \sqrt{3})
\]

### Ratio of Intensities \(I_P : I_Q\)
\[
\text{Ratio} = \frac{I_P}{I_Q} = \frac{3A^2}{A^2(2 + \sqrt{3})} = \frac{3}{2 + \sqrt{3}}
\]
Rationalizing the denominator:
\[
\frac{3}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{3(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{6 - 3\sqrt{3}}{4 - 3} = 6 - 3\sqrt{3}
\]

Thus, the ratio of intensities is:
\[
\boxed{6 - 3\sqrt{3}}
\]

Let me know if you'd like further details! 😊

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1. What is the significance of the phase difference in wave interference?  
2. How does the amplitude affect the intensity of light?  
3. What are the boundary conditions for constructive and destructive interference?  
4. How can the ratio of intensities be experimentally measured?  
5. What happens if the amplitudes \(A_1\) and \(A_2\) are unequal?  

**Tip**: Always remember that the cosine of the phase difference determines the nature of interference (constructive or destructive).

Learn how to calculate the ratio of intensities at points P and Q in a Young's double slit experiment with phase differences of λ/6 and λ/12. The solution involves understanding wave interference, phase difference, and intensity formulas. Suitable for senior high school physics students.

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