In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? (The central bright fringe is zeroth one).

Answers

Answer 1

Answer:

The wavelength of the light passing through the slit is 214 nm.

What is the wavelength?

The wavelength is the distance between identical points in the adjacent cycles of a waveform.

Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.

The wavelength of the light passing through the slit is calculated as given below.

$\lambda = \frac {sd}{mr}$

$\lambda = \frac {10* 10^(-2) * 3.00 * 10^(-5)}{7 * 2}$

$\lambda = 2.14 * 10^(-7)\;\rm m$

$\lambda = 214 \;\rm nm$

Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.

To know more about the wavelength, follow the link given below.

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Answer 2

Answer:

To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.

By definition in the principle of superposition, light interference is defined as

$Y = (m\lambda R)/(d)$

Where,

d = Separation of the two slits

$\lambda = Wavelength$

R = Distance from slit to screen

m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.

Y = Distance from central spot to fringe.

Re-arrange the equation to find \lambda we have that

$\lambda = (Yd)/(mR)$

Our values are gives according the problem as,

$Y = 10*10^(-2)m$

$d = 3*10^(-5)$

m = 7 (The seventh bright fringe)

R = 2m

$\lambda = ((10*10^(-2))(3*10^(-5)))/(7*2)$

$\lambda = 214nm$

Therefore the wavelength of the light passing through the slits is 214nm

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1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:
1. 5 cm = ________ mm
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Answers

$5cm = 50mm$
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$4.378kg = 4378g$
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a weight is suspended from the ceiling by a spring (k = 20 ln/in) and is connected to the floor by a dashpot producing viscous damping. The damping force is 10 lb when the velocity of the dashpot plunger is 20 in/sec. The weight and plunger have W = 12 lb. What will be the frequency of the damped vibrations?

Answers

Answer:

The frequency of the damped vibrations is 3.82 Hz.

Explanation:

Given that,

Spring constant = 20 lb/in

Damping force = 10 lb

Velocity = 20 in/sec

Weight = 12 lb

We need to calculate the damping constant

Using formula of damping force

$b* v=F_(d)$

$b=(F_(d))/(v)$

Put the value into the formula

$b =(10)/(20)$

$b=0.5\ lb-sec/in$

$b=0.5*12 =6\ lb-sec/ft$

We need to calculate the frequency

Using formula of angular frequency

$\omga=\sqrt{\omega_(0)^2-((b)/(2m))^2}$

$\omega=\sqrt{(k)/(m)-((b)/(2m))^2}$

Put the value into the formula

$\omega=\sqrt{(20*12*32)/(12)-((6*32)/(2*12))^2}$

$\omega=24\ rad/s$

We need to calculate the frequency of the damped vibrations

Using formula of frequency

$f=(\omega)/(2\pi)$

Put the value into the formula

$f=(24)/(2\pi)$

$f=3.82\ Hz$

Hence, The frequency of the damped vibrations is 3.82 Hz.

An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is : (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line

Answers

Answer:

2.72 m

Explanation:

wavelength of sound λ = velocity / frequency

= 340 / 1200

= .2833 m

Distance of point of first constructive interference

= λ D / d ( D is distance of the screen and d is distance between source of sound.

Here D = 12.5 m

d = 1.3 m

λ D / d= ( .2833 x 12.5) / 1.3

= 2.72 m

Distance of point of first constructive interference = 2.72 m

Final answer:

The wavelength of the produced sound is approximately 0.29 m. Constructive interference occurs when the path difference between the two waves is a multiple of this wavelength, allowing you to calculate the distance between the central maximum and first maximum loud position.

Explanation:

For part (a) of the question, we need to calculate the wavelength of the sound wave. The wave speed (v) is given by the multiplication of frequency (f) and wavelength (λ). The speed of sound in air is approximately 343 m/s and given that the frequency produced by the function generator is 1200 Hz, the wavelength can be calculated using the formula λ = v / f = 343 / 1200 ≈ 0.29 m.

For part (b) the distance between the central maximum (loud) position and the first maximum along this line requires understanding of sound wave interference and constructive interference. For constructive interference to occur, the path difference between the two waves needs to be a multiple of the wavelength. Thus, in the first constructive interference position (first maximum loud position), the path difference equals one wavelength (0.29m). Since the student is walking 12.5 m away and parallel to the line between the speakers (which is the hypotenuse of a right triangle stakeout, with one side being 0.65m), we can use Pythagorean theorem to find out the distance.

Learn more about Sound Waves Interference here:

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Which of the following describes the net force acting on an object?The sum of all forces acting on an object
The gravitational force minus any contact forces acting on an object
The difference between the normal force and the gravitational force acting on an object
The sum of all the forces acting on an object in the same direction

Answers

The sum of all forces acting on an object in the same direction is described for the net force acting on an object.

What is a Net force?

When the forces are acting in the same direction of movement of the object it can be said as sum of the two individual forces will be equal to the "Net Force" .

The net force is the combined force of all individual forces acting on an object.
If the object with the forces in the opposite direction, then the net force will not be equal to the sum of the forces.

Example : If two forces (2 kids pushing in the same direction to move the object big box) act on an object (big box) in the same direction, then the net force is equal to the sum of the two forces. If the kids pushed in the opposite direction, the net force will not occur.

Hence, Option D is the correct answer.

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Answer:

The sum of all the forces acting on an object in the same direction.

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mirror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object