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Gemiola [76]
3 years ago
5

A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that

the first completely dark fringes occur on the screen at distances of ±8.24 mm from the central bright fringe, and that the central bright fringe has an intensity of 2.00 W/m2 at its center. (a) How wide is the slit?
Physics
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

a = 2.275 10⁻⁴ m

Explanation:

This is a diffraction problem that is described by the equation

         a sin θ = m λ

         

The first dark minimum occurs for m = 1

         a = λ  / sin θ

The angle can be found by trigonometry,

       tan θ = y / x

       θ = tan⁻¹ y / x

Let's reduce the magnitudes to the SI system

        y = 8.24 mm = 8.24 10⁻³ m

        λ  = 625 nm = 625 10⁻⁹ m

       θ = tan⁻¹ 8.24 10⁻³ / 3.00

       θ = 0.002747 rad

We calculate

       a = 625 10⁻⁹ / sin 0.002747

       a = 2.275 10⁻⁴ m

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2.464 cm above the water surface

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Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

weight of the block = 0.78 * 11.2^3  gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

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6 0
3 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

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