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crimeas [40]
3 years ago
12

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.350 mm with the screen 39.0 cm away fr

om the slit, when light of wavelength 570 nm is used.
(a) Find the slit width.
(b) Calculate the angle θ of the first diffraction minimum.
Physics
1 answer:
rjkz [21]3 years ago
5 0

Answer: a) the width is 2.54 mm

b) the angle is 0.012°

Explanation:

we have the equation:

λ = zw/(L*m)

Where is the distance for the middle of the screen, L is the distance to the screen, w is the widht of the slit and λ is the wavelength.

now we can isolate z and get:

z = λL*m/w

and the distance between the 5 minimum and the first one is 0.035mm

then:

0.035mm = λL*5/w - λL*1/w = λL*4/w

now we can solve it for w.

0.35mm = 570nm*39.0cm*4/w

now, we have all diferent units, lets use nanometers.

1cm = 1x10^-2 m

1mm = 1x10^-3 m

1nm = 1x10^-9m

then: 1 cm = 1x10^7 nm

         1 mm = 1x10^6 nm

then we have:

0.35x10^6 nm = 570nm*39.0x10^7 nm*4/w

w = (570nm*39.0x10^7 nm*4)/0.35x10^6 nm =(570nm*39.0*10*4)/0.35

w = 2540571.4 nm

it is a bigg number, let's write it in milimeters:

w = (2540571.4/10^6) mm = 2.54 mm

the first minimum can be obtained by the equation w*sinθ = mλ by using m = 1

then we have:

2540571.4 nm*sinθ = 570 nm

sinθ = 570/2540571.4 = 0.000224

θ = asin(0.00022) = 0.012°

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In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

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