The Setup
When Light Comes in Sideways
Imagine a standard Young's Double Slit Experiment (YDSE). Usually, we shine a parallel beam of light perfectly perpendicular to the plane of the slits. In that ideal scenario, the light reaches both slits at the exact same time, meaning the initial phase difference is zero. But what happens if we tilt the light source?
In this problem, the incident parallel beam is coming in at an angle α from above the horizontal axis. Because of this tilt, the wavefront is no longer parallel to the slit plane. The upper ray hits the upper slit first, while the lower ray has to travel a slightly extra distance to reach the lower slit. This extra distance creates an initial path difference before the light even enters the slits!
Using simple geometry, if the distance between the slits is d, the extra distance traveled by the lower ray is Δxinitial=dsinα. Since the angle α is extremely small, we can safely use the small angle approximation: Δxinitial≈dα.
The Master Equation
Adding Path Differences
Now, let's look at what happens after the light passes through the slits. For any point P on the screen located at a distance y above the central axis, the ray from the lower slit has to travel further than the ray from the upper slit. This introduces a final path difference given by the standard YDSE formula: Δxfinal=dsinθ≈dDy.
Here is the crucial catch: because the light is incident from above, the lower ray travels further before the slits AND after the slits to reach point P. Therefore, these two path differences compound. The total path difference at point P is the sum of both:
ΔxP=Δxinitial+Δxfinal=dα+dDy
Option A
The Center is No Longer the Center
Let's test Option A, which asks about the interference at the geometric center O (where y=0).
First, we must convert the given angle α=π0.36∘ into radians. This is a classic trap! The small angle approximation sinα≈α only works if α is in radians.
α=π0.36×180π=2×10−3 rad
At point O, the final path difference is zero. So, the total path difference is just the initial one:
ΔxO=dα=(3×10−4 m)×(2×10−3 rad)=6×10−7 m
Notice that 6×10−7 m is exactly 600 nm, which is our wavelength λ. Since the path difference is exactly 1⋅λ (an integer multiple), we get constructive interference at O. Option A claims it is destructive, so Option A is incorrect.
Option B
Does the Pattern Shrink?
Option B suggests that the fringe spacing depends on α. Let's recall the formula for fringe width:
Look closely at the formula—there is no α! The oblique incidence merely adds a constant initial path difference to every point on the screen. This causes the entire interference pattern to shift downwards, but the distance between consecutive bright fringes remains completely unchanged. Option B is incorrect.
Option C & D
The Fate of Point P
Now let's evaluate point P, located at y=11.0 mm=11×10−3 m. We plug our values into the master equation:
ΔxP=(6×10−7 m)+(3×10−4 m)(1 m11×10−3 m)
ΔxP=6×10−7+33×10−7=39×10−7 m
To determine the type of interference, we divide the total path difference by the wavelength λ to find the order n:
n=λΔxP=6×10−739×10−7=6.5
Since n=6.5 is a half-integer (6+21), the waves arrive exactly out of phase, resulting in destructive interference. Option C is absolutely correct!
Finally, for Option D, if α=0, the initial path difference vanishes. The total path difference at P would just be 33×10−7 m. Dividing by λ gives n=5.5, which is again a half-integer, meaning destructive interference. Option D claims constructive, so it is incorrect.
The Grand Takeaway: Oblique incidence doesn't change the nature of the fringes; it just shifts the entire pattern. The new central maximum (where Δx=0) will now form below the geometric center at y=−Dα.
