Solution to Question 15
1. Geometric Setup and Phase Analysis
We consider a cylindrical screen of radius $R$ illuminated by two plane coherent beams, A and B.
Let us define the coordinate system and direction vectors:
- The center of the cylinder is at the origin $(0,0)$.
- Beam A propagates horizontally along the positive x-axis. Its wave vector is $\vec{k}_A = k \hat{i}$.
- Beam B propagates at an angle $\theta$ relative to Beam A. Based on the geometry where A and B subtend an angle $\theta$ and point P is at an angle $\phi$ “outside” the region between them (as shown in the figure), the angle between the direction of B and the position vector of P will be $(\phi + \theta)$.
- Point P is located on the screen at an angular position $\phi$ from the direction of Beam A. Its position vector is $\vec{r}$.
Figure 1: Schematic of beams A and B incident on the cylindrical screen. $\theta$ is the angle between beams, and $\phi$ is the angular position of P from beam A.
2. Calculation of Optical Phase
The phase of a plane wave at a point defined by position vector $\vec{r}$ is given by $\Phi = \vec{k} \cdot \vec{r}$.
For Beam A:
The angle between the wave vector $\vec{k}_A$ and the radius vector $\vec{r}$ to point P is $\phi$. Thus:
$$ \Phi_A = k R \cos(\phi) $$
For Beam B:
Beam B propagates at an angle $\theta$ with respect to Beam A. From the geometry, the angle between the wave vector $\vec{k}_B$ and the radius vector $\vec{r}$ is $(\phi + \theta)$. Thus:
$$ \Phi_B = k R \cos(\phi + \theta) $$
3. Path Difference and Interference Condition
The phase difference $\Delta \Phi$ at point P is:
$$ \Delta \Phi = \Phi_A – \Phi_B = k R [\cos\phi – \cos(\phi + \theta)] $$
To find the distance between adjacent fringes (fringe width), we determine the change in angular position $d\phi$ corresponding to a change in phase difference of $2\pi$ (or path difference of $\lambda$).
Differentiating $\Delta \Phi$ with respect to $\phi$:
$$ \frac{d(\Delta \Phi)}{d\phi} = k R \left[ -\sin\phi – (-\sin(\phi + \theta)) \right] $$
$$ \frac{d(\Delta \Phi)}{d\phi} = k R \left[ \sin(\phi + \theta) – \sin\phi \right] $$
Using the trigonometric identity $\sin C – \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$:
$$ \sin(\phi + \theta) – \sin\phi = 2 \cos\left(\frac{\phi + \theta + \phi}{2}\right) \sin\left(\frac{\phi + \theta – \phi}{2}\right) $$
$$ = 2 \cos\left(\phi + \frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) $$
4. Calculating Fringe Width
The condition for the separation between adjacent fringes is $|d(\Delta \Phi)| = 2\pi$. Let the linear fringe width be $w = R d\phi$.
$$ \left| \frac{d(\Delta \Phi)}{d\phi} \right| d\phi = 2\pi $$
$$ k R \left[ 2 \cos\left(\phi + \frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) \right] d\phi = 2\pi $$
Substituting $k = \frac{2\pi}{\lambda}$:
$$ \frac{2\pi}{\lambda} (R d\phi) \left[ 2 \cos\left(\phi + \frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right) \right] = 2\pi $$
Solving for the fringe width $w = R d\phi$:
$$ w = \frac{\lambda}{2 \cos\left(\phi + \frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right)} $$
The distance between adjacent interference fringes near point P is:
$$ \frac{\lambda}{2 \cos\left(\phi + \frac{\theta}{2}\right) \sin\left(\frac{\theta}{2}\right)} $$
We consider a cylindrical screen of radius $R$ illuminated by two plane coherent beams, A and B. Let us define the coordinate system and direction vectors:
- The center of the cylinder is at the origin $(0,0)$.
- Beam A propagates horizontally along the positive x-axis. Its wave vector is $\vec{k}_A = k \hat{i}$.
- Beam B propagates at an angle $\theta$ relative to Beam A. Based on the geometry where A and B subtend an angle $\theta$ and point P is at an angle $\phi$ “outside” the region between them (as shown in the figure), the angle between the direction of B and the position vector of P will be $(\phi + \theta)$.
- Point P is located on the screen at an angular position $\phi$ from the direction of Beam A. Its position vector is $\vec{r}$.
Figure 1: Schematic of beams A and B incident on the cylindrical screen. $\theta$ is the angle between beams, and $\phi$ is the angular position of P from beam A.
The phase of a plane wave at a point defined by position vector $\vec{r}$ is given by $\Phi = \vec{k} \cdot \vec{r}$.
For Beam A:
The angle between the wave vector $\vec{k}_A$ and the radius vector $\vec{r}$ to point P is $\phi$. Thus:
For Beam B:
Beam B propagates at an angle $\theta$ with respect to Beam A. From the geometry, the angle between the wave vector $\vec{k}_B$ and the radius vector $\vec{r}$ is $(\phi + \theta)$. Thus:
The phase difference $\Delta \Phi$ at point P is:
To find the distance between adjacent fringes (fringe width), we determine the change in angular position $d\phi$ corresponding to a change in phase difference of $2\pi$ (or path difference of $\lambda$).
Differentiating $\Delta \Phi$ with respect to $\phi$:
Using the trigonometric identity $\sin C – \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$:
The condition for the separation between adjacent fringes is $|d(\Delta \Phi)| = 2\pi$. Let the linear fringe width be $w = R d\phi$.
Substituting $k = \frac{2\pi}{\lambda}$:
Solving for the fringe width $w = R d\phi$: