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xz_007 [3.2K]
3 years ago
5

In a Young's double-slit experiment, 610-nm-wavelength light is sent through the slits. The intensity at an angle of 2.95° from

the central bright fringe is 85% of the maximum intensity on the screen. What is the spacing between the slits?
Physics
1 answer:
Sliva [168]3 years ago
7 0

Answer:

spacing between the slits is 405.32043 ×10^{-9}  m

Explanation:

Given data

wavelength = 610 nm

angle = 2.95°

central bright fringe = 85%

to find out

spacing between the slits

solution

we know that spacing between slit is

I = 4I_{0} × cos²∅/2

so

I/4I_{0}  = cos²∅/2

here I/4I_{0} is 85 % = 0.85

so

0.85 = cos²∅/2

cos∅/2 = √0.85

∅ = 2 ×cos^{-1} 0.921954

∅  = 45.56°

∅  = 45.56° ×π/180 = 0.7949 rad

and we know that here

∅  = 2π d sinθ / wavelength

so

d = ∅× wavelength /  ( 2π  sinθ )

put all value

d = 0.795 × 610×10^{-9} / ( 2π  sin2.95 )

d = 405.32043 ×10^{-9}  m

spacing between the slits is 405.32043 ×10^{-9}  m

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Yuliya22 [10]

Answer:

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c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
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Answer:

a. cosθ b. E.A

Explanation:

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b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
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LVULAN
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For this case we have that by definition, the kinetic energy is given by the following formula:

k= \frac {1} {2} * m * v ^ 2

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

m = 100 \ kg\\v = 9 \frac {m} {s}

Substituting the values we have:

k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050

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Answer:

Option A

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Answer:

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4 0
3 years ago
a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti
PolarNik [594]

Answer:

B=0.2T

Explanation:

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B=0.4/2=0.2T

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3 years ago
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