Fringe Shift in YDSE with Thin Film

1. The Key Idea: Optical Path Difference

In a standard YDSE, the central fringe is bright because the waves from slits S1 and S2 travel equal geometric distances to the center of the screen, arriving in phase.

When we insert a transparent sheet of thickness $t$ and refractive index $\mu$ in front of one slit (say, S1), the light wave passing through it slows down. While it travels a geometric distance of $t$ through the sheet, it takes more time than the wave in the air. This delay is equivalent to the wave having traveled a longer path in a vacuum. This "equivalent vacuum distance" is called the optical path.

The optical path for a distance $t$ in a medium of refractive index $\mu$ is $\mu t$.

The other wave (from S2) travels the same thickness $t$ in the air (or vacuum, where $\mu \approx 1$). So, the sheet introduces an additional optical path difference between the two waves.

Additional Path Difference, $\Delta x_{sheet}=(\text{Optical Path in sheet})-(\text{Optical Path in air})$
$\Delta x_{sheet}=\mu t-1\cdot t=(\mu -1)t$

This simple expression, $(\mu -1)t$, is the fundamental cause of everything that follows.

To see a clear explanation of this concept, please watch the following video segment.

2. Calculating the Fringe Shift

Because of this new path difference, the point at the center of the screen (y=0) is no longer a point of zero total path difference. It will no longer be the central maximum.

The central maximum is defined as the point where the total path difference is zero. The entire fringe pattern must shift on the screen to a new position where the geometric path difference perfectly cancels out the optical path difference introduced by the sheet.

Let's say the pattern shifts by a distance $S$. At this new position, the geometric path difference is $\Delta x_{geom}=\frac{Sd}{D}$ (from our last lesson, with $y=S$). For this to be the new central maximum, this geometric path difference must exactly compensate for the path difference from the sheet.

$\Delta x_{total}=\Delta x_{geom}-\Delta x_{sheet}=0$
$\frac{Sd}{D}=(\mu -1)t$

Solving for the fringe shift, $S$, we get our main formula:

$S=\frac{D(\mu -1)t}{d}$

Important notes:

• The shift $S$ is in the direction of the slit where the sheet was placed. If you place it in front of S1 (the upper slit), the pattern shifts up. If you place it in front of S2 (the lower slit), the pattern shifts down.
• The fringe width, $\beta =\frac{\lambda D}{d}$, does not change. The sheet only moves the entire pattern; it doesn't stretch or compress it.

The following video provides a step-by-step derivation of this fringe shift formula.

3. Shift in Terms of Fringe Width

For problem-solving, it's often useful to express the shift in terms of the number of fringe widths. This tells us how many fringes the central maximum has moved.

We have the shift: $S=\frac{D(\mu -1)t}{d}$
And the fringe width: $\beta =\frac{\lambda D}{d}$

Dividing $S$ by $\beta$, we get the number of fringes shifted, $N$:

$N=\frac{S}{\beta }=\frac{\frac{D(\mu -1)t}{d}}{\frac{\lambda D}{d}}$

$N=\frac{(\mu -1)t}{\lambda }$

This is an extremely useful formula for NEET problems that ask questions like "the central maximum shifts to the position of the 5th bright fringe." In that case, $N=5$.

4. Problem Solving for NEET

As promised, let's focus on applying these formulas to exam-style questions. The following resource provides excellent solved examples that cover the most common problem variations.

Let's break down the strategy for a few of these problem types:

• Problem Type 1: Calculating the shift (Example 4)

  • Given: Thickness ($t$), refractive index ($\mu$), and wavelength ($\lambda$).
  • Goal: Find the shift in terms of number of fringes.
  • Method: Use the formula $N=\frac{(\mu -1)t}{\lambda }$ directly. The answer will be "N fringes upward/downward".

• Problem Type 2: Finding an unknown parameter (Example 1)

  • Given: The central maximum shifts by $n$ fringe widths.
  • Goal: Find the thickness $t$ of the sheet.
  • Method: Use the same formula, but rearranged: $n=\frac{(\mu -1)t}{\lambda }⟹t=\frac{n\lambda }{\mu -1}$.

• Problem Type 3: Intensity at the old center (Example 3)

  • Given: The intensity at the original central position remains unchanged (i.e., it's still a maximum).
  • Goal: Find the thickness $t$ (or $x\lambda$ in the problem).
  • Method: For the intensity to be a maximum, the path difference introduced by the slab must be an integral multiple of the wavelength. So, set $(\mu -1)t=n\lambda$, where $n$ is an integer (1, 2, 3...). The question asks for the minimum non-zero thickness, so you'd use $n=1$. In the example, they find $x=2n$, and the smallest non-zero integer choice is $x=2$.

Test your understanding!

The image below shows a YDSE setup where two different glass slabs are placed in front of the two slits. What is the path difference at the central point O on the screen?

Show answer

This is a great problem to test your understanding of how path differences add up.

1. Path difference from the top slab (S1): The slab has thickness $t$ and refractive index ${\mu }_1=1.5$. It introduces an optical path difference of $\Delta x_1=({\mu }_1-1)t=(1.5-1)t=0.5t$. This path is "longer" or "slower".

2. Path difference from the bottom slab (S2): The slab has thickness $t$ and refractive index ${\mu }_2=1.7$. It introduces an optical path difference of $\Delta x_2=({\mu }_2-1)t=(1.7-1)t=0.7t$. This path is even "slower".

3. Net Path Difference at O: At the central point O, the geometric path difference is zero. The total path difference is therefore just the difference between the optical path differences introduced by the two slabs.
   $\Delta x_{net}=\Delta x_2-\Delta x_1$
   $\Delta x_{net}=0.7t-0.5t=0.2t$

The net path difference at the central point O is $0.2t$. The fringe pattern will shift towards the slab that causes a greater delay, so the new central maximum will be located below the point O.

Conclusion

Today we've extended our knowledge of YDSE to include the important case of introducing a transparent sheet. By understanding the concept of optical path, we were able to derive the formulas for the resulting fringe shift.

Key Takeaways:

• A transparent sheet of thickness $t$ and refractive index $\mu$ introduces an extra optical path difference of $\Delta x=(\mu -1)t$.
• This causes the entire fringe pattern to shift by a distance $S=\frac{D(\mu -1)t}{d}$ towards the slit with the sheet.
• The fringe width $\beta$ does not change.
• The shift, measured in the number of fringes, is $N=\frac{(\mu -1)t}{\lambda }$.

Preview of the Next Lesson:
The concept of optical path difference is not just for YDSE. It's the key to understanding another beautiful interference phenomenon: the iridescent colors you see in soap bubbles and thin oil slicks on water. In our next lesson, we will explore thin-film interference, where we'll see how light waves reflecting from the top and bottom surfaces of a thin film interfere with each other. The principles we learned today will be directly applicable.