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gavmur [86]
3 years ago
13

If 730-nm and 640-nm light passes through two slits 0.61 mm apart, how far apart are the second-order fringes for these two wave

lengths on a screen 1.1 m away?
Physics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

X2 - X2'=3.24×10⁻⁴m

Explanation:

Given Data

λ1=730nm=

λ2=640nm

d=0.61 mm

D=1.1 m

Distance of the second bright fringe from the central fringe is given by

Xn = = D x n x2 λ / d

X2 = 2 D λ1 / d

X2' = 2 D λ2 / d

Separation between the second bright fringes of the two wavelengths ( the separation between the second order fringes ) is ,

X2 - X2' = 2 D ( λ1 - λ2 ) / d

X2 - X2'= {2 x 1. 1 (730×10^-9 - 640×10^-9) }/ 0.61 x 10 ^-3

X2 - X2'=3.24×10⁻⁴m

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Gas like oxygen, nitrogen etc

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

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7 0
3 years ago
a man uses a 900 watt electric iron to press his clothes for an average of 4 hours a week for five weeks if the cost of an elect
ser-zykov [4K]

Answer:

54 Kobo

Explanation:

Units of <u>electricity</u> are measured in kilowatt hours (kWh).

Given information:

  • 900 watt electric iron
  • Appliance usage = 4 hours a week for 5 weeks
  • Unit cost of electricity = 3 Kobo per kWh

<h3><u>Step 1</u></h3>

Convert the wattage of the electric iron from watts to kilowatts.

1000 watts (W) = 1 kilowatt (kW)

⇒ 900 watts = 1 ÷ 1000 = 0.9 kilowatts

This means that the power consumption of the electric iron is 0.9 kW per hour of use.

<h3><u>Step 2</u></h3>

Total hours spent pressing clothes:

= 4 hours per week for 5 weeks

= 4 × 5

= 20 hours

<u>Total power consumption</u>:

= number of kW × number of hours

= 0.9 × 20

= 18 kWh

<h3><u>Step 3</u></h3>

To find the <u>total cost</u>, multiply the total kWh by the cost per kWh:

⇒ Cost = 18 × 3 = 54 Kobo

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What happened to the speed of light if it travels from air into glass?
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Predict using Boyle's law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa.
kobusy [5.1K]

The volume of the balloon will halve

Explanation:

Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,

pV=const.

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

p_1 V_1 = p_2 V_2

And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:

p_1 = 101 kPa is the initial pressure at sea level (the atmospheric pressure)

V_1 is the initial volume

p_2 = 202 kPa is the final pressure

V_2 is the final volume

Substituting into the equation, we find:

V_2 = \frac{p_1 V_1}{p_2}=\frac{(101)V_1}{202}=\frac{V_1}{2}

Which means that the volume of the balloon will halve.

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6 0
4 years ago
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