Physics, 12.08.2020 08:01 mprjug6

A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen. Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.

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The wavelength of this light is approximately $427\; \rm nm$ ( $4.27\times 10^{-7}\; \rm m$ .)The distance between the first and central maxima is approximately $7.40\; \rm cm$ (about twice the distance between the first dark fringe and the central maximum.)

Explanation:

Wavelength

Convert all lengths to meters:

Separation of the two slits: $0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m$ .Distance between the first dark fringe and the center of the screen: $3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m$ .

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be $\theta$ .

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be $\lambda / 2$ (one-half the wavelength of the light.)

Therefore:

$\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}$ .

On the other hand:

$\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}$ .

Because the cotangent of $\theta$ is very close to zero,

$\cos \theta \approx \cot \theta \approx 0.00711538$ .

$\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538$ .

$\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}$ .

Distance between two adjacent maxima

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

The path difference required for the central maximum is

.The path difference required for the first maximum is $\lambda$ .The path difference required for the second maximum is $2\,\lambda$ .

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

$\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}$ .

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is $3.70\; \rm cm$ when the path difference is $\lambda / 2$ . The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: $2 \times 3.70\; \rm cm = 7.40\; \rm cm$ .

A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.

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A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0....