which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C

Answer:

Explanation:

Convert all lengths to meters:

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)

Therefore:

[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].

On the other hand:

[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].

Because the cotangent of [tex]\theta[/tex] is very close to zero,

[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].

[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].

[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].

Answer:

a) Connect a series resistance of 8,47 ohms

b)14,16 [A]

c) r = 10,96 ohms

Explanation:

My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower

In America

P = V*I

1700 (w) = 120*I

I = 1700/120 [A]

I = 14,16 [A]        current needed for the blower

In Europe

120 (v)  (the drop of voltage I need) when a current of 14,16 passes through to series  resistor is

V = I*R          120 = 14,16* R         R = 8,47 ohms

c) P = I*r²

1700 (w) = 14,16 (A) * r²

r² = 120,06

r = 10,96 ohms

Answer:

a) for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

b) therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

c) Given the torque is 71.0% of its maximum value; Ф  = 45.24⁰ ≈ 45⁰

Explanation:

Given that; Diameter is 8.40 cm,

Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m

Number of turns (N) = 17

Current in the loop (I) = 3.20 A

Magnetic field (B) = 0.610 T

Let the angle between the loop's area vector A and the magnetic field B be

Now. the area of the loop is;

A = πR²

A = 3.14 ( 0.042 )²

A =  0.005539 m²

Torque on the loop (t) = NIABsinФ

t = 17 × 3.20 ×0.005539 × 0.610 × sinФ

t = 0.1838sinФ N.m

for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

Given the torque is 71.0% of its maximum value

t = 0.71 × tmax

t = 0.71 × 0.1838

t = 0.1305

Now

0.1305 N.m =  0.1838 sinФ N.m

sinФ = 0.1305 / 0.1838

sinФ = 0.71001

Ф = sin⁻¹ 0.71001

Ф  = 45.24⁰ ≈ 45⁰

Answer:

The amplitude of the resultant wave is 12.93 cm.

Explanation:

The amplitude of resultant of two waves, y₁ and y₂, is given as;

Y = y₁ + y₂

Let y₁ = A sin(kx - ωt)

Since the wave is out phase by φ, y₂ is given as;

y₂ = A sin(kx - ωt + φ)

Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )

Given;

phase difference, φ = 45°

Amplitude, A = 7.00 cm

Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )

Y = 12.93 cm

Therefore, the amplitude of the resultant wave is 12.93 cm.

Answer:

The length of the tube is 85 cm

Explanation:

Given;

speed of sound, v = 340 m/s

first harmonic of open-closed tube is given by;

N----->A , L= λ/₄

λ₁ = 4L

v = Fλ

F = v / λ

F₁ = v/4L

Second harmonic of open-closed tube is given by;

L = N-----N + N-----A, L = (³/₄)λ

[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]

Third harmonic of open-closed tube is given by;

L = N------N + N-----N + N-----A, L = (⁵/₄)λ

[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]

The difference between second harmonic and first harmonic;

[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]

The difference between third harmonic and second harmonic;

[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]

Thus, the difference between successive harmonic of open-closed tube is

v / 2L.

[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]

Therefore, the length of the tube is 85 cm

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

Answer:

The period of motion is 0.5 second.

Explanation:

Given;

extension of the spring, x = 3.8 cm = 0.038 m

mass of the object, m = 13 g = 0.013 kg

Determine the force constant of the spring, k;

F = kx

k = F / x

k = mg / x

k = (0.013 x 9.8) / 0.038

k = 3.353 N/m

When the object is replaced with a block of mass 20 g, the period of motion is calculated as;

[tex]T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second[/tex]

Therefore, the period of motion is 0.5 second.

Answer:

Resistance of each bulb = 360 ohms

Explanation:

Let each bulb have a resistance r .

Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.

[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]

[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]

R(equivalent) = r/15

Now, As per Ohms Law :

V = I * R(equivalent)

120 V = 5 A * r/15

r = 360 ohms

Answer:

a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².

Explanation:

a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.

τ = Frsinθ

= 185 N × 1.73 m × sin90°

= 320.05 Nm

So, the torque τ = 320.05 Nm

b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m

α = angular acceleration of grindstone

τ = Iα

α = τ/I = τ/(MR²/2) = 2τ/MR²

substituting the values of the variables, we have

α = 2τ/MR²

= 2 × 320.05 Nm/[76 kg × (1.73 m)²]

= 640.1 Nm/227.4604 kgm²

= 2.814 rad/s²

So, the angular acceleration α = 2.814 rad/s²

c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.

So  τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm

The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm

Since τ'' = Iα

α' = τ''/I where α' = its new angular acceleration

α' = 2τ/MR²

= 2 × 319.75 Nm/[76 kg × (1.73 m)²]

= 639.5 Nm/227.4604 kgm²

= 2.811 rad/s²

So, the angular acceleration α' = 2.811 rad/s²

Answer:

200cm

Explanation:

Answer:

100cm

Explanation:

Using

F= ( N/2L)(√T/u)

F1 will now be (0.5*2)( √600/0.015)

=> L( wavelength)= 200/2cm = 100cm

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]

So it would normally be written with a negative signs in front indicating a divergent lens.

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Answer:

The mass of the object is 3.08 kg.

Explanation:

The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.

Let the mass of the object = m.

The coefficient of kinetic friction, n = 0.42

Therefore,  

Force, F = n × mg

12.7 = 0.42 × 9.8 × m

m = 3.08 kg

The mass of the object is 3.08 kg.

Answer:

Explanation:

Current in the same direction

 When current flow through to parallel conductors of a given length, when the current flows in the same direction

1. A force of attraction between the wires occurs and this tends to draw the wires inward

2. A magnetic field in the same direction is produced.

Current in opposite direction

when the current is in opposite direction

1. Force of repulsion between the two wires occurs, draws the wire outward

2. A magnetic field in opposite direction occurs

Answer:

Explanation:

Using the lens formula

1//f = 1/u+1/v

f is the focal length of the lens

u is the object distance

v is the image distance

For convex lens

The focal length of a convex lens is positive and the image distance can either be negative or positive.

Given f = 20cm and u = 10cm

1/v = 1/f - 1/u

1/v = 1/20-1/10

1/v = (1-2)/20

1/V = -1/20

v = -20/1

v = -20 cm

Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image

For concave lens

The focal length of a concave lens is negative and the image distance is negative.

Given f = -20cm and u = 10cm

1/v = 1/f - 1/u

1/v = -1/20-1/10

1/v = (-1-2)/20

1/V = -3/20

v = -20/3

v = -6.67 cm

Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image

Explanation:

According to Ohms Law :

V = I * R

(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms

Also,

[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]

(B)

[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]

Drift speed is missing. It is given as;

1.7 × 10^(-5) m/s

A) R = 0.0293 ohms

B) ρ = 1.589 × 10^(-8)

C) n = 8.8 × 10^(28) electrons

This is about finding, resistance and resistivity.

Length; L = 5.8 m

Diameter; d = 2mm = 0.002 m

Radius; r = d/2 = 0.001 m

Voltage; V = 22 mv = 0.022 V

Current; I = 750 mA = 0.75 A

Area; A = πr² = 0.001²π

Drift speed; v_d = 1.7 × 10^(-5) m/s

R = V/I

R = 0.022/0.75

R = 0.0293 ohms

ρ = RA/L

ρ = (0.0293 × 0.001²π)/5.8

ρ = 1.589 × 10^(-8)

J = n•e•v_d

Where;

J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²

e is charge on an electron = 1.6 × 10^(-19) C

v_d = 1.7 × 10^(-5) m/s

n is number of free electrons per unit volume

Thus;

238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))

238732.44 = (2.72 × 10^(-24))n

n = 238732.44/(2.72 × 10^(-24))

n = 8.8 × 10^(28)

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Answer:

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force

Explanation:

galvanometer is an instrument that can detect and measure small current in an electrical circuit.

If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.

The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

Answer:

The energy of one of its photons is 1.391 x 10⁻¹⁸ J

Explanation:

Given;

wavelength of the UVC light, λ = 142.9 nm = 142.9 x 10⁻⁹ m

The energy of one photon of the UVC light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f is frequency of the light

f = c / λ

where;

c is speed of light = 3 x 10⁸ m/s

λ  is wavelength

substitute in the value of f into the main equation;

E = hf

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{142.9*10^{-9}} \\\\E = 1.391*10^{-18} \ J[/tex]

Therefore, the energy of one of its photons is 1.391 x 10⁻¹⁸ J

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

Answer:

Explanation:

We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.

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Answer:

[tex]\boxed{\sf 65.3 \ inches}[/tex]

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

Answer: her rotational kinetic energy increases

Answer:

Acceleration of the body is:

[tex]a=0.27\,\,m/s^2[/tex]

Explanation:

Use Newton's second Law to solve for the acceleration:

[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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Answer:

The polarity of the magnetic field changes

Explanation:

This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)

Complete Question

A toroidal solenoid has 590 turns, cross-sectional area 6.20 cm^2 , and mean radius 5.00 cm .

Part A. Calculate  the coil's self-inductance.

Part B. If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self-induced emf in the coil.

Part C. The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?

Answer:

Part A  

       [tex]L = 0.000863 \ H[/tex]

Part B  

       [tex]\epsilon = 0.863 \ V[/tex]

Part C

    From terminal a to terminal b

Explanation:

From the question we are told that

      The  number of turns is  [tex]N = 590 \ turns[/tex]

      The cross-sectional area is  [tex]A = 6.20 cm^2 = 6.20 *10^{-4} \ m[/tex]

      The  radius is [tex]r = 5.0 \ cm = 0.05 \ m[/tex]

Generally the coils self -inductance is mathematically represented as

              [tex]L = \frac{ \mu_o N^2 A }{2 \pi * r }[/tex]

Where [tex]\mu_o[/tex] is the permeability of  free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting values

             [tex]L = \frac{ 4\pi * 10^{-7} * 590^2 6.20 *10^{-4} }{2 \pi * 0.05 }[/tex]

             [tex]L = \frac{ 2 * 10^{-7} * 590^2 6.20 *10^{-4} }{ 0.05 }[/tex]

             [tex]L = 0.000863 \ H[/tex]

Considering the Part B

      Initial current is [tex]I_1 = 5.00 \ A[/tex]

      Current at time t is [tex]I_t = 3.0 \ A[/tex]

       The  time taken is  [tex]\Delta t = 3.00 ms = 0.003 \ s[/tex]

The self-induced emf is mathematically evaluated as

          [tex]\epsilon = L * \frac{\Delta I}{ \Delta t }[/tex]          

=>         [tex]\epsilon = L * \frac{ I_1 - I_t }{ \Delta t }[/tex]

substituting values

             [tex]\epsilon = 0.000863 * \frac{ 5- 2 }{ 0.003 }[/tex]  

             [tex]\epsilon = 0.863 \ V[/tex]

The direction of the induced emf is  from a to b because according to Lenz's law the induced emf moves in the same direction as the current

This question involves the concepts of the self-inductance, induced emf, and Lenz's Law

A. The coil's self-inductance is "0.863 mH".

B. The self-induced emf in the coil is "0.58 volts".

C. The direction of the induced emf is "from b to a".

A.

The self-inductance of the coil is given by the following formula:

[tex]L=\frac{\mu_oN^2A}{2\pi r}[/tex]

where,

L = self-inductance = ?

[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ N/A²

N = No. of turns = 590

A = Cross-sectional area = 6.2 cm² = 6.2 x 10⁻⁴ m²

r = radius = 5 cm = 0.05 m

Therefore,

[tex]L=\frac{(4\pi\ x\ 10^{-7}\ N/A^2)(590)^2(6.2\ x\ 10^{-4}\ m^2)}{2\pi(0.05\ m)}[/tex]

L = 0.863 x 10⁻³ H = 0.863 mH

B.

The self-induced emf is given by the following formula:

[tex]E=L\frac{\Delta I}{\Delta t}\\\\[/tex]

where,

E = self-induced emf = ?

ΔI = change in current = 2 A

Δt = change in time = 3 ms = 0.003 s

Therefore,

[tex]E=(0.000863\ H)\frac{2\ A}{0.003\ s}[/tex]

E = 0.58 volts

C.

According to Lenz's Law, the direction of the induced emf always opposes the change in flux that causes it. Hence, the direction of the induced emf will be from b to a.

Learn more about Lenz's Law here:

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Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m