Question

item 1
a very fine wire $7.35 \times 10^{-3}$ mm in diameter is placed between two flat glass plates as in (figure 1). light whose wavelength in air is 520 nm falls (and is viewed) perpendicular to the plates and a series of bright and dark bands is seen
figure
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part a
how far apart are the dark fringes if the glass plates are each 23.5 cm long?
express your answer to three significant figures and include the appropriate units.
$\delta x = $ value units
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Explanation:

Step1: Convert units to meters

Diameter of wire \( d = 7.35\times10^{-3}\space mm = 7.35\times10^{-6}\space m \)
Wavelength \( \lambda = 520\space nm = 520\times10^{-9}\space m \)
Length of plates \( L = 23.5\space cm = 0.235\space m \)

Step2: Use the formula for fringe separation in air wedge

The formula for the distance between dark fringes \( \Delta x \) in an air wedge is \( \Delta x=\frac{\lambda L}{2d} \) (derived from the condition for dark fringes in air wedge interference, where the path difference and phase change are considered).

Substitute the values:
\( \Delta x=\frac{(520\times 10^{-9}\space m)(0.235\space m)}{2\times(7.35\times 10^{-6}\space m)} \)

First, calculate the numerator: \( (520\times 10^{-9})(0.235)=520\times0.235\times 10^{-9}=122.2\times 10^{-9}\space m^2 \)

Then, calculate the denominator: \( 2\times7.35\times 10^{-6}=14.7\times 10^{-6}\space m \)

Now, divide numerator by denominator: \( \frac{122.2\times 10^{-9}}{14.7\times 10^{-6}}=\frac{122.2}{14.7}\times 10^{-3}\space m \approx 8.31\times 10^{-3}\space m = 8.31\space mm \) (Wait, let's recalculate the division: \( 122.2\div14.7\approx8.31 \), so \( 8.31\times 10^{-3}\space m = 8.31\space mm \)? Wait, no, wait: \( 10^{-9}/10^{-6}=10^{-3} \), so yes. Wait, but let's check the formula again. Wait, the air wedge formula: the condition for dark fringes is \( 2t = m\lambda \), where \( t \) is the thickness of the air wedge at a distance \( x \) from the contact point, and \( t=\frac{d}{L}x \) (from similar triangles, since the wire is at the end, so \( \frac{t}{x}=\frac{d}{L} \), so \( t = \frac{d}{L}x \)). Substituting into \( 2t = m\lambda \), we get \( 2\frac{d}{L}x = m\lambda \), so \( x=\frac{m\lambda L}{2d} \). Then the separation between consecutive dark fringes ( \( \Delta m = 1 \)) is \( \Delta x=\frac{\lambda L}{2d} \). So that formula is correct.

Wait, let's recalculate the numbers:

\( \lambda = 520\space nm = 520\times 10^{-9}\space m = 5.2\times 10^{-7}\space m \)

\( L = 0.235\space m \)

\( d = 7.35\times 10^{-3}\space mm = 7.35\times 10^{-6}\space m \)

So \( \Delta x=\frac{(5.2\times 10^{-7})(0.235)}{2\times(7.35\times 10^{-6})} \)

Calculate numerator: \( 5.2\times0.235\times 10^{-7}=1.222\times 10^{-7}\space m^2 \)

Denominator: \( 2\times7.35\times 10^{-6}=1.47\times 10^{-5}\space m \)

Now divide: \( \frac{1.222\times 10^{-7}}{1.47\times 10^{-5}}=\frac{1.222}{1.47}\times 10^{-2}\space m \approx 0.00831\space m = 8.31\space mm \)? Wait, no: \( 10^{-7}/10^{-5}=10^{-2} \), so \( 1.222/1.47\approx0.831 \), so \( 0.831\times 10^{-2}\space m = 8.31\times 10^{-3}\space m = 8.31\space mm \). Wait, but that seems large? Wait, no, the wire diameter is \( 7.35\times 10^{-3}\space mm = 7.35\space \mu m \), length is 23.5 cm = 235 mm. Wavelength is 520 nm = 0.52 \mu m. So let's use \mu m units to make it easier.

\( d = 7.35\space \mu m \)

\( \lambda = 0.52\space \mu m \)

\( L = 235\space mm \)

Then \( \Delta x=\frac{\lambda L}{2d}=\frac{0.52\space \mu m\times235\space mm}{2\times7.35\space \mu m} \)

Convert mm to \mu m: 235 mm = 235000 \mu m

So \( \Delta x=\frac{0.52\times235000}{2\times7.35}\space \mu m=\frac{0.52\times235000}{14.7}\space \mu m \)

Calculate 0.52*235000 = 122200

122200 /14.7 ≈ 8312.93 \mu m = 8.31\space mm. Ah, so that's correct. So the fringe separation is approximately 8.31 mm? Wait, but let's check with the initial numbers:

Wait, \( d = 7.35\times 10^{-3}\space mm = 7.35\times 10^{-6}\space m \)

\( L = 0.235\space m \)

\( \lambda = 520\times 10^{-9}\space m \)

So \( \Delta x=\frac{520\times 10^{-9}\times0.235}{2\times7.35\times 10^{-6}} \)

Calculate numerator: 520e-9 * 0.235 = 520*0.235 e-9 = 122.2 e-9 m²

Denominator: 2*7.35e-6 = 14.7e-6 m

Divide: 122.2e-9 /14.7e-6 = (122.2/14.7) * 1e-3 m ≈ 8.31 * 1e-3 m = 8.31 mm. Yes, that's correct. So the distance between dark fringes is approximately 8.31 mm. Wait, but let's check significant figures. The given values: diameter is 7.35×10⁻³ mm (three sig figs), wavelength 520 nm (three sig figs), length 23.5 cm (three sig figs). So the answer should have three sig figs. So 8.31 mm? Wait, but let's recalculate the division:

122.2 /14.7: 14.7*8 = 117.6, 122.2-117.6=4.6, 4.6/14.7≈0.313, so 8.313, so 8.31 mm (three sig figs). So the answer is 8.31 mm, or 8.31×10⁻³ m.

Answer:

\( \Delta x = \boxed{8.31\space mm} \) (or \( 8.31\times 10^{-3}\space m \))