PHYSICS COLLEGE

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

Answer 1

Answer:

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) = 514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ / d ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

= 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations : brainly.com/question/4449144

Answer 2

Answer:

Answer:

$1.7* 10^(-3)$ m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

$w = (D\lambda )/(d)$

$w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))$

$w = 1.7* 10^(-3)$ m

Related Questions

Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 V. However, the battery delivers only 2.5 V. Therefore, we decide to charge two equal value capacitors in parallel from the 2.5V battery and then switch the capacitors in series with the heart during the 1ms pulse. What is the minimum value of the capacitances required so the output pulse amplitude remains between 4.9 V and 5.0 V throughout its 1ms duration
Select the correct answer.If the coefficient of static friction is 0.35 and the normal force is 80 newtons, what is the maximum frictional force of the surface acting on the object? O A. B. C. OD. OE 9.8 newtons 28 newtons 80 newtons 23 newtons 35 newtons
(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)
A train moves with a uniform velocity of 36km/hr 10sec. calculate the distance travelled
A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.20 ?

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.(a) Find the value of the normal force on the block.

(b) Find the magnitude of force F~2 that is acting on the block

(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .

Answers

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

Answers

A diverging lens always produces a virtual erect image.

The general lens formula is given as;

$(1)/(F) = (1)/(U) + (1)/(V)$

Where;

U = object distance
V = image distance
F = focal length of the lens

A lens can be converging or diverging.

A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.

The image produced by a diverging lens is always virtual and upright.

Thus, we can conclude that a diverging lens always produces a virtual erect image.

Learn more here:brainly.com/question/11788630

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

A car is decelerating at the rate of 2 km/s square. If its initial speed is 66 km/s, how long will it take the car to come to a complete stop?

Answers

Answer:

It will take 33 seconds to stop the car.

Explanation:

Using the first equation of kinematics we have

$v=u+at$

where

'v' is final speed of object

'u' is initial speed of object

'a' is acceleration of object

't' is time of acceleration of object

Now since it is given that $a=-2km/s^(2)$ since acceleration is negative and $u=66km/s$

We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get

$0=66-2* t\n\n\therefore t=(66)/(2)=33seconds$

Why is it important to select a coordinate system when studying motion?

Answers

Motion is detected when an object changes its position with respect to a reference point. Coordinate system is basically used to represent motion. A coordinate system uses numbers or coordinates which represent position of the reference points on a two-dimensional or three-dimensional space. The trajectory of a point or line can be studied on a coordinate system which describes various aspects of motion like velocity, acceleration, distance, displacement etc. Coordinate system is important because it helps to choose a starting point and the direction (which will be positive).

. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.

You should obtain e/m = 2V/(B^2)(r^2)

3. The magnetic field on the axis of a circular current loop a distance z away is given by

B = mu I R^2 / 2(R^2 + z^2)^ (3/2)

where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:

|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/R

where B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Answers

Answer:

Explanation:

Magnetic field creates a force perpendicular to a moving charge in its field which is equal to Bev where B is magnetic field , e is amount of charge on the moving charge and v is the velocity of charge particle .

This force provides centripetal force for creation of circular motion. If r be the radius of the circular path

Bev = mv² / r

r = mv / Be

2 ) If an electron is accelerated by an electric field created by potential difference V then electric field

= V / d where d is distance between two points having potential difference v .

force on charged particle

electric field x charge

= V /d x e

work done by field

= force x distance

= V /d x e x d

V e

This is equal to kinetic energy created

V e = 1/2 mv²

= 1/2 m (r²B²e² / m² )

V = r²B²e/ 2 m

e / m = 2 V/ r²B²

3 )

B = $(\mu* I* R^2)/(2(R^2+Z^2)^(3)/(2) )$

In Helmholtz coils , distance between coil is equal to R so Z = R/2

B = $(\mu* I* R^2)/(2(R^2+(R^2)/(4) )^(3)/(2) )$

For N turns of coil and total field due to two coils

B = $(\mu* I* N)/(R*((5)/(4))^(3)/(2) )$

= $(\mu* I* N)/(R)* ((4)/(5))^(3)/(2)$

= 9.0 x 10^-7 NI/R

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

$Tan \theta = (y)/(D)$

where D is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy = |y1max (λ1) − y1max (λ2)|

Δy = $|(D\lambda _1)/(d) - (D\lambda _2)/(d) |$

Δy = $D |(d/20)/(d) - (d/15)/(d) |$

Δy = $D |(1)/(20) - (1)/(15) |$

Δy = $4.90 |(1)/(20)- (1)/(15) |$

Δy = 81.7mm

The separation between these maxima is 81.7 mm

Δy = |y₂max (λ1) − y₂max (λ2)|

Δy = $D|(2(d/20))/(d) - (5(d/15))/(2d) |$

Δy = $4.90|(1)/(10) - (1)/(6) |$

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

Learn more about Wave Interference here:

brainly.com/question/16622714

#SPJ12

Other Questions

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes. If instead of a single wire we use a coil of thin Nichrome wire containing 23 turns, what does the ammeter read?

Two capacitors give an equivalent capacitance of 9.42 pF when connected in parallel and an equivalent capacitance of 1.68 pF when connected in series. What is the capacitance of each capacitor?

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.