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3 years ago

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The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small c

ompared to the distance from the slits to the screen, the distance between adjacent fringes near the center of the interference pattern _______.

1 answer:

Elodia [21]3 years ago

4 0

Answer:

the distance between adjacent fringes is increased by a factor o 2

Explanation:

To find how the distance between fringes is modified you can use the following formula for the calculation of the distance between fringes:

$\Delta y=\frac{\lambda D}{d}$

D: distance to the screen

d: distance between slits

λ: wavelength of the light

if d is decreased by a factor of 2, that is d'=1/2d, you have:

$\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y$

hence, the distance between adjacent fringes is increased by a factor o 2

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Explain how fossil fuel use in the US makes us vulnerable to the demands of foreign countries.

daser333 [38]

Well the US doesn't naturally have a lot of fossil fuels so we rely on other countries who do have fossil fuels. So if we aren't that friendly with Saudia Arabia (they have a bunch of oil) then they can hike up the prices or cut us off from their oil supply completely which would cripple the US severly. So with that being said, we NEED other countries, and the big question is when will they not NEED us. At any point in which they feel that they don't need us they can just cut us off, which would be bad.

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3 years ago

Read 2 more answers

A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh

maksim [4K]

Answer: $16.234^{\circ}C$

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water( $m_w$ )=600 gm

temperature of water =15

specific heat of water $=4.186 j/g-^{\circ}C$

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.) $=mgh=6\times 9.81\times 10=588.6 J$

Heat supplied $=m_wc(\Delta T)$

$H.E.=600\times 4.186\times (T-16)$

$588.6=2511.6\times (T-16)$

T-16=0.234

$T=16.234^{\circ}C$

This is not an efficient way of heating water as there is only $0.234^{\circ}C$ increase in temperature.

5 0

3 years ago

Ferromagnetic materials-

QveST [7]

Answer:

materials which exhibit a spontaneous net magnetization at the atomic level, even in the absence of an external magnetic field.

Explanation:

When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. This effect is known as Faraday's Law of Magnetic Induction. However, materials can react quite differently to the presence of an external magnetic field. This reaction is dependent on a number of factors, such as the atomic and molecular structure of the material, and the net magnetic field associated with the atoms. The magnetic moments associated with atoms have three origins. These are the electron motion, the change in motion caused by an external magnetic field, and the spin of the electrons.

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3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

$C = \frac{\epsilon_{o} A }{d}$

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

$Q = 1.245 \times 10^{-12}$

3 0

3 years ago

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

3 years ago