Solution for Question 16
Part (a): The Interference Pattern
The interference occurs between spherical wavefronts from the point source $S$ and planar wavefronts from the parallel beam.
Consider a point $P$ on the screen at a radial distance $r$ from the center (the point directly below $S$).
- The path length of the ray from the parallel beam to reach the screen is constant everywhere (assuming normal incidence), equal to the perpendicular distance $l$.
- The path length of the ray from the point source $S$ to point $P$ is the hypotenuse $\sqrt{l^2 + r^2}$.
The path difference $\Delta x$ at radius $r$ is:
$$ \Delta x = \sqrt{l^2 + r^2} – l $$
Since the locus of points having a constant path difference depends only on $r$ (distance from the center), the pattern consists of concentric circles.
At the center ($r=0$), $\Delta x = 0$. This corresponds to a constructive interference maximum (since the sources are in phase). Therefore, the pattern consists of circular fringes with the central one being a bright spot.
Part (b): Spacing between fringes
For the $n^\text{th}$ bright fringe, the path difference must be an integer multiple of the wavelength:
$$
\sqrt{l^2 + r_n^2} – l = n\lambda
$$
$$
\sqrt{l^2 + r_n^2} = l + n\lambda
$$
Squaring both sides:
$$
l^2 + r_n^2 = (l + n\lambda)^2 = l^2 + 2ln\lambda + n^2\lambda^2
$$
$$
r_n^2 = 2ln\lambda + n^2\lambda^2
$$
Assuming $l \gg n\lambda$, the term $n^2\lambda^2$ is negligible compared to $2ln\lambda$.:
$$
r_n = \sqrt{2ln\lambda + n^2\lambda^2} \approx \sqrt{2\lambda l n} \quad (\text{for large } l)
$$
The spacing between the $n^\text{th}$ and $(n-1)^\text{th}$ fringe is $\Delta r = r_n – r_{n-1}$. Using the approximation $r_n \approx \sqrt{2\lambda l}\sqrt{n}$:
$$
\Delta r = \sqrt{2\lambda l}\sqrt{n} – \sqrt{2\lambda l}\sqrt{n-1}
$$
Expression for spacing:
$$
\text{Spacing} \approx \sqrt{2\lambda l} (\sqrt{n} – \sqrt{n-1})
$$
Part (a): The Interference Pattern
The interference occurs between spherical wavefronts from the point source $S$ and planar wavefronts from the parallel beam.
Consider a point $P$ on the screen at a radial distance $r$ from the center (the point directly below $S$).
- The path length of the ray from the parallel beam to reach the screen is constant everywhere (assuming normal incidence), equal to the perpendicular distance $l$.
- The path length of the ray from the point source $S$ to point $P$ is the hypotenuse $\sqrt{l^2 + r^2}$.
The path difference $\Delta x$ at radius $r$ is: $$ \Delta x = \sqrt{l^2 + r^2} – l $$
Since the locus of points having a constant path difference depends only on $r$ (distance from the center), the pattern consists of concentric circles.
At the center ($r=0$), $\Delta x = 0$. This corresponds to a constructive interference maximum (since the sources are in phase). Therefore, the pattern consists of circular fringes with the central one being a bright spot.
Part (b): Spacing between fringes
For the $n^\text{th}$ bright fringe, the path difference must be an integer multiple of the wavelength:
$$ \sqrt{l^2 + r_n^2} – l = n\lambda $$ $$ \sqrt{l^2 + r_n^2} = l + n\lambda $$Squaring both sides:
$$ l^2 + r_n^2 = (l + n\lambda)^2 = l^2 + 2ln\lambda + n^2\lambda^2 $$ $$ r_n^2 = 2ln\lambda + n^2\lambda^2 $$Assuming $l \gg n\lambda$, the term $n^2\lambda^2$ is negligible compared to $2ln\lambda$.:
$$ r_n = \sqrt{2ln\lambda + n^2\lambda^2} \approx \sqrt{2\lambda l n} \quad (\text{for large } l) $$The spacing between the $n^\text{th}$ and $(n-1)^\text{th}$ fringe is $\Delta r = r_n – r_{n-1}$. Using the approximation $r_n \approx \sqrt{2\lambda l}\sqrt{n}$:
$$ \Delta r = \sqrt{2\lambda l}\sqrt{n} – \sqrt{2\lambda l}\sqrt{n-1} $$