The bar magnet is pushed toward the center of a wire loop. Looking down from the top view (would appear the magnet is coming up toward the observer); Which is true? A. There is no induced current in the loop B. There is a counterclockwise induced current in the loop C. There is not enough information to correctly answer the question D. There is a clockwisee induced current in the loop

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:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

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:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

[tex]\beta = \frac{\lambda D}{d}[/tex]

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

Answer:

Fad = 28.8 kN

Fbd = 16.4 kN

Fcd = 28.1 kN

Explanation:

First, find the length of each cable.

AD = √((2 m)² + (0.5 m)² + (2.5 m)²)

AD = √10.5 m

AD ≈ 3.24 m

BD = √((1.5 m)² + (1 m)² + (2.5 m)²)

BD = √9.5 m

BD ≈ 3.08 m

CD = √((1 m)² + (1 m)² + (2.5 m)²)

CD = √8.25 m

CD ≈ 2.87 m

Next, use similar triangles to find the x, y, and z components of each tension force.

Fadx = 2/3.24 Fad = 0.617 Fad

Fady = 0.5/3.24 Fad = 0.154 Fad

Fadz = 2.5/3.24 Fad = 0.772 Fad

Fbdx = 1.5/3.08 Fbd = 0.487 Fbd

Fbdy = 1/3.08 Fbd = 0.324 Fbd

Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd

Fcdx = 1/2.87 Fcd = 0.348 Fcd

Fcdy = 1/2.87 Fcd = 0.348 Fcd

Fcdz = 2.5/2.87 Fcd = 0.870 Fcd

Now sum the forces in the x, y, and z directions:

∑Fx = ma

-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0

∑Fy = ma

-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0

∑Fz = ma

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.

-0.463 Fad + 0.811 Fbd = 0

0.811 Fbd = 0.463 Fad

Fbd = 0.571 Fad

Substitute into either of the first two equations:

-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0

-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0

-0.339 Fad + 0.348 Fcd = 0

0.348 Fcd = 0.339 Fad

Fcd = 0.975 Fad

Now substituting into the third equation:

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0

60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0

60 kN − 2.083 Fad = 0

Fad = 28.8 kN

Solving for the other two tension forces:

Fbd = 0.571 Fad = 16.4 kN

Fcd = 0.975 Fad = 28.1 kN

Answer:

Tensions of:

DA = 28.81 KN

DB = 16.45 KN

DC = 28.07 KN

Explanation:

see attached

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:

See attached file

Explanation:

Answer: her rotational kinetic energy increases

Answer:

Explanation:i think this would help u

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.

hope this helps

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Answer:

Explanation:

Bohr's atomic model was replaced by the  modern atomic model because of its limitations, which included :

(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )

(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).

(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

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:

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:

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:

Φ = 36.84 × 10^(-20) J

Explanation:

In the photoelectric effect, the energy of the incoming photon is usually used in part to extract the photoelectron from the material (work function) and then the rest is converted into kinetic energy of the photoelectron which is given by the formula;

K_max = hf - Φ

where;

hf represents the energy of the incoming photon

h is the Planck's constant

f is the light frequency

Φ is the work function of the material

K_max is the maximum kinetic energy of the photoelectrons.

From the question, we are told that no current flows unless the wavelength is less than 540 nm. This means that when the wavelength has this value, the maximum kinetic energy of the photoelectrons is zero i.e K_max = 0. Thus the energy of the incoming photons is just enough to extract the photoelectrons from the material.

Thus,

hf - Φ = 0

hf = Φ - - - (1)

We are given the wavelength as;

λ = 540 nm = 540 × 10^(-9) m

Now, let's find the frequency of the light by using the relationship between frequency and wavelength. The equation is;

f = c/λ

Where c is speed of light = 3 × 10^(8) m/s

f = (3 × 10^(8))/(540 × 10^(-9))

f = 5.56 × 10^(14) Hz

Thus, from equation 1,we can now find the work function;

Φ = hf

h is Planck's constant and has a value of 6.626 × 10^(-34) J.s

Thus;

Φ = 6.626 × 10^(-34) × 5.56 × 10^(14)

Φ = 36.84 × 10^(-20) J

The correct answer is C. The inner planets are also called terrestrial planets.

Explanation:

Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".

Answer:

The answer is c.) The inner planets are also called terrestrial planets.

Explanation:

Answer:

A The father should sit closer to the pivot.

C The longer wrench makes the job easier because less force is needed when there is more distance from the pivot.

A As far from the head of the hammer as possible because this will maximize torque.

D at the opposite side of the seesaw towards the middle

:) gl

Explanation:

If a father and his son want to play on a seesaw then to balance the torque of the seesaw the father should sit near the pivot as he had more weight as compared to his son, while the son should sit a little farther from the pivot point as compared to his father.

Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.

Mechanical advantage = output force(load) /input force (effort)

As given in the problem statement If a father and his son wish to play on a seesaw,

The father should sit close to the pivot because he weighs more than his son, and the son should sit a little farther away from the pivot point than his father. This will help balance the torque of the seesaw.

Thus, the father should sit near the pivot on the one side and the son should sit a little farther from the pivot of a seesaw on the other side.

Learn more about Mechanical advantages, here

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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)   I = I₀/2, b)  I = I₀/2 cos² θ

Explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

          i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

          I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

                  I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

                I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0          I = Io / 2

θ = 90º      I = 0

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

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:

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:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

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)

Answer:

The volume will decrease by a factor of 10/51.

Explanation:

Hello,

In this case, since both moles and temperature remain constant, we can use the Boyle's law that relates the volume and pressure as an inversely proportional relationship:

[tex]P_1V_1=P_2V_2[/tex]

Thus, since the pressure increases by a factor of 5.1 (statement), we have:

[tex]P_2=5.1P_1[/tex]

Thus, the final volume is:

[tex]V_2=\frac{P_1V_1}{P_2} =\frac{P_1V_1}{5.1P_1}\\\\V_2=\frac{10}{51}V_1[/tex]

It means that the volume will decrease by a factor of 10/51.

Regards.

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

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

f = c / λ

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

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].