Answer:
43.16°
Explanation:
λ = Wavelength = 1.4×10⁻¹⁰ m
θ₁ = 20°
n can be any integer
d = distance between the two slits
Since for the first bright fringe, n₁ = 1
n₂ = 2 for second order line
The relation between the distance of the slits and the angle through which it is passed is:
dsinθ=nλ
As d and λ are constant
![\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_1%5Clambda%7D%7Bsin%20%5Ctheta_1%7D%3D%5Cfrac%7Bn_2%5Clambda%7D%7Bsin%20%5Ctheta_2%7D%5C%5C%5CRightarrow%20%5Cfrac%7B1%7D%7Bsin20%7D%3D%5Cfrac%7B2%7D%7Bsin%5Ctheta_2%7D%5C%5C%5CRightarrow%20sin%5Ctheta_2%3D%5Cfrac%7B2%7D%7B%5Cfrac%7B1%7D%7Bsin20%7D%7D%5C%5C%5CRightarrow%20%5Ctheta_2%3Dsin%5E%7B-1%7D%7B%5Cfrac%7B2%7D%7B%5Cfrac%7B1%7D%7Bsin20%7D%7D%7D%5C%5C%5CRightarrow%20%5Ctheta_2%3D43.16%5E%7B%5Ccirc%7D)
∴ Angle by which the second order line appear is 43.16°
Answer:
The current in the wire = 39.53 A.
Explanation:
The given cross -sectional area of the wire = 1.35
.
The number of electrons flowing in the wire at any point is
= 5.41
electrons.
The charge of an electron is known to be = ![1.6 \times 10^{-19} \hspace{0.1cm} Coulombs](https://tex.z-dn.net/?f=1.6%20%5Ctimes%2010%5E%7B-19%7D%20%5Chspace%7B0.1cm%7D%20Coulombs)
The electronics flow across the wire for a time of, t = 2.19 seconds.
The current in the wire = (number of electrons) × ![\frac{charge \hspace{0.1cm} of \hspace{0.1cm} an \hspace{0.1cm} electron}{time}](https://tex.z-dn.net/?f=%5Cfrac%7Bcharge%20%5Chspace%7B0.1cm%7D%20of%20%5Chspace%7B0.1cm%7D%20an%20%5Chspace%7B0.1cm%7D%20electron%7D%7Btime%7D)
= 5.41
× ![\frac{1.6 \times 10^{-19}}{2.19}](https://tex.z-dn.net/?f=%5Cfrac%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D%7B2.19%7D)
= 39.53 A.
The formula for current is actually, i =
.
The (number of electrons )×(charge of an electron) will equal the total charge flowing through the wire in the given time = 2.19 seconds.
The current in the wire = 39.53 A.
W=Fd
W=22,000N*(2,000m)
W=44,000,000J
Answer:
E. d and O
Explanation:
"Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings".
According to Huygens’s principle, "for each element of the wavefront in the slit emits wavelets. These are like rays that start out in phase and head in all directions. (Each ray is perpendicular to the wavefront of a wavelet.) Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel".
The destructive interference for a single slit is given by:
![d sin \theta = m\lambda , m=1,-1,2,-2,3,...](https://tex.z-dn.net/?f=d%20sin%20%5Ctheta%20%3D%20m%5Clambda%20%2C%20m%3D1%2C-1%2C2%2C-2%2C3%2C...)
Where
d is the slit width
is the light's wavelength
is the angle relative to the original direction of the light
m is the order od the minimum
I represent the intensity
When the intensity and the wavelength are incident normally the angular as we can see on the expression above the angular separation just depends of the distance d and the wavelength O.