To find the distance separating successive dark fringes in the interference pattern created by the wedge-shaped film of air between the two glass slides, we can use the principles of thin film interference.
We start with the formula for the location of dark fringes in an air wedge:
\[ y_n = \frac{(n + 0.5) \lambda L}{t} \]
where:
- \(y_n\) is the distance from the contact point at which the \(n\)-th dark fringe occurs,
- \(n\) is the order of the dark fringe (0, 1, 2, ...),
- \(\lambda\) is the wavelength of the light in the medium (in this case, it's in air),
- \(L\) is the length of the air wedge (which is the length of the slides, 10 cm = 0.1 m),
- \(t\) is the thickness of the wedge at the point where the measurement is taken.
First, we convert the wavelength to the compatible units: \[ \lambda = 0.402 , \mu m = 0.402 \times 10^{-6} , m \]
Now, using the total thickness \(t = 65 , \mu m = 65 \times 10^{-6} , m\), we see that the thickness \(t\) remains constant as 65 μm at the far end of the slides.
The distance between successive dark fringes \( \Delta y \) can be determined by finding the spacing between the positions of dark fringes:
\[ \Delta y = \frac{\lambda L}{t} \]
Next, we input the relevant values:
\[ \Delta y = \frac{0.402 \times 10^{-6} , m \times 0.1 , m}{65 \times 10^{-6} , m} \]
Calculating this gives:
\[ \Delta y = \frac{0.402 \times 10^{-7}}{65 \times 10^{-6}} \] \[ \Delta y = \frac{0.402 \times 10^{-7}}{65 \times 10^{-6}} = \frac{0.402}{65} \times 10^{-1} m \] \[ \Delta y \approx 6.18 \times 10^{-3} \times 10^{-1} m \] \[ \Delta y \approx 6.18 \times 10^{-4} m \] \[ \Delta y \approx 0.618 , mm \]
Thus, the distance separating successive dark fringes when viewed from above is:
\[ \boxed{0.618 , mm} \]
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