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3 years ago

14

How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th

e line density of the grating were doubled?

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Answer:

a) the distance between the interference fringes is reduced by half

b) the distance between stripes is doubled

Explanation:

Interference experiments constructive interference is described by the expression

d sin θ = m λ

let's use trigonometry to find the distance between the interference fringes

tan θ= y / L

dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles

tan θ = sin θ / cos θ = sin θ

we substitute

sin θ = y / L

d y / L = m λ

y = m λ / d L

a) it asks us when the screen doubles its distance

L ’= 2 L

subtitute in the equation

y ’= m λ / (d 2L)

y ’=( m λ / d L) /2

y ’= y / 2

the distance between the interference fringes is reduced by half

b) the density of the network doubles

if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half

d ’= d / 2

we substitute

y ’= m λ (L d / 2)

y ’= m λ / (L d) 2

y ’= y 2

the distance between stripes is doubled

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A girl playing tug-of-war with her dog pulls the dog a distance of 8.0m by exerting a force at an angle of 18° with the horizont

AnnZ [28]

Answer:

25 N

Explanation:

Work is a product of force and perpendicular distance moved.

W=Fd where F is force exerted and d is perpendicular distance.

However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as $dcos\theta$

Therefore, $W=Fdcos\theta$

Making F the subject of the formula then

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$F=\frac {190}{8cos18^{\circ}}\approx 25N$

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3 years ago

A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re

White raven [17]

A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law.

$I= \frac{V}{R}$

Where:
I = current (A)
V = Voltage (V) (emf)
R = Resitance (Ω)

Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
$V = IR$

However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
$\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}$

$\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}$

$\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059$
$\frac{1}{R_{T}} =0.0613$

Get the reciprocal of both sides and divide:

$R_{T} = \frac{1}{0.0613} =16.32$

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
$V = IR$
$V = (1.8)(16.32)$
$V = 29.376V$

The emf of the battery is 29.376V

B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{65}$
$I= 0.45A$

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{25}$
$I= 1.18A$
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
$I= \frac{V}{R}$
$I= \frac{29.376}{170}$
$I= 0.17A$

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.

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Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

$h=v_{0} t+\frac{gt^{2}}{2}$

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

$h=\frac{gt^{2}}{2}$

We know the height h of the spaceship hovering, and the gravity of Venus is $g=8.87\frac{m}{s^{2}}$ . Substituting this values in the equation $h=\frac{gt^{2}}{2}$ :

$17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}$

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

$t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s$

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth $g=9.81\frac{m}{s^{2}}$ .

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