Power = (voltage) x (current)
100w = 220v x current
Current = 100w / 220v = <u>0.455 Ampere</u> (rounded)
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
![y_n= \frac{n \lambda D}{a}](https://tex.z-dn.net/?f=y_n%3D%20%5Cfrac%7Bn%20%5Clambda%20D%7D%7Ba%7D)
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
![D=37.0 cm=0.37 m](https://tex.z-dn.net/?f=D%3D37.0%20cm%3D0.37%20m)
![\lambda=530 nm=5.3 \cdot 10^{-7} m](https://tex.z-dn.net/?f=%5Clambda%3D530%20nm%3D5.3%20%5Ccdot%2010%5E%7B-7%7D%20m)
while the distance between the first and the fifth minima is
![y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m](https://tex.z-dn.net/?f=y_5-y_1%20%3D%200.500%20mm%3D0.5%20%5Ccdot%2010%5E%7B-3%7D%20m)
(2)
If we use the formula to rewrite
![y_5, y_1](https://tex.z-dn.net/?f=y_5%2C%20y_1)
, eq.(2) becomes
![\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%20%5Clambda%20D%7D%7Ba%7D%20-%20%5Cfrac%7B1%20%5Clambda%20D%7D%7Ba%7D%20%3D%5Cfrac%7B4%20%5Clambda%20D%7D%7Ba%7D%3D%200.5%20%5Ccdot%2010%5E%7B-3%7D%20m%20%20)
Which we can solve to find a, the width of the slit:
Look it up it's really easy :) I don't really remember all of this. I haven't learned about this stuff since like 7th grade
The best answer among the following choices would be A) or the first option Scientists want to share measurements data that they can understand.