Two glass microscope slides, of length 10 cm, are in contact at one end and separated by a sheet of paper at the other end, which creates a wedge-shaped film of air. The thickness of the paper is t = 65 μm, and the glass has index of refraction n = 1.5. Light with wavelength 0.402 μm shines vertically down on the slides. In the resulting interference pattern, what is the distance separating successive dark fringes when viewed from above?

Answer 1

To find the distance separating successive dark fringes in the interference pattern, we can use the principles of thin film interference. Since the air wedge formed by the separation of the glass slides due to the thickness of the paper is very thin compared to the length of the slides, we can derive the distance between dark fringes using the following relationship derived from the geometry of the wedge and interference conditions.

1. Wavelength in the Air Film: Since light is passing through a film of air and the refractive index of air is approximately 1, we will use the wavelength of light in air directly. The wavelength in air (\(\lambda\)) is given as \(0.402 , \mu m\).

2. Thickness of the Air Wedge: The thickness of the paper creates a triangular wedge of air which has a thickness \( t = 65 , \mu m\).

3. Phase Change: For constructive interference at dark fringes, a \( \frac{1}{2} \lambda \) phase shift occurs upon reflection from the denser medium (glass). Thus for dark fringes, the condition for destructive interference at points where the path length difference (\( \Delta \)) is an odd multiple of half-wavelengths (for air, and considering \( \Delta = m\lambda\) where \(m\) is an integer): \[ 2t = (m + 0.5)\lambda \] For successive dark fringes, we can set up an equation based on the wedge thickness at a distance \(x\) from the point of contact.

4. Wedge Geometry Relation: The thickness \( t \) of the air wedge at a distance \( x \) can be approximated using the small angle approximation since the thickness linearly varies from \(0\) to \(65 , \mu m\) over \(10 , cm\): \[ t(x) = \frac{65 , \mu m}{10 , cm} \cdot x = 0.0065 x \quad \text{(in mm)} \]

Thus, now using the dark fringe condition: \[ 2t(x) = (m + 0.5)\lambda \] Substituting for \(t(x)\): \[ 2 \cdot (0.0065 x) = (m + 0.5) \cdot 0.402 , \mu m \]

Solving for \( x \) gives: \[ x = \frac{(m + 0.5) \cdot 0.402 , \mu m}{2 \cdot 0.0065} \]

5. Distance Between Successive Dark Fringes: By taking the difference \( \Delta x \) between successive values of \( m \) (i.e., \( 0 \rightarrow 1 \)): \[ \Delta x = \frac{((m+1) + 0.5) \cdot 0.402 - (m + 0.5) \cdot 0.402}{2 \cdot 0.0065} \] \[ \Delta x = \frac{0.402 , \mu m}{2 \cdot 0.0065} \] This simplifies to: \[ \Delta x = \frac{0.402 \times 10^{-6} m}{0.013} \approx 0.0310 , m \approx 31.0 , mm \]

Therefore, the distance separating successive dark fringes when viewed from above is approximately 31.0 mm.