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4 years ago

What is meant by constructive and destructive interference

1 answer:

6 0

In waveform physics, constructive and destructive interference describes the way two different waves interact with one another. If two waves are aligned such that the peaks of both are in the same spot, the waves experience constructive interference - the two add together and become a single, more powerful wave. The opposite happens if the peak of one meets the trough of the other - the two cancel each other out, and the magnitude of the combined wave is lesser than the individual waves. There are also other, more complex versions of interference if they don’t align cleanly, but the most common descriptions of interference are the ones described above.

Hope this helps!

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In a diffraction grating experiment, light of 600 nm wavelength produces a first-order maximum 0.350 mm from the central maximum

NARA [144]

Answer:

497.143 nm.

Explanation:

Diffraction grating experiment is actually done by passing light through diffraction glasses, the passage of the light causes some patterns which can be seen on the screen. This is because light is a wave and it can spread.

The solution to the question is through the use of the formula in the equation (1) below;

Sin θ = m × λ. ---------------------------------(1).

Where m takes values from 0, 1, 2, ...(that is the diffraction grating principal maxima).

Also, m × λ = dc/ B -------------------------------------------(2).

We are to find the second wavelength, therefore;

λ2 =( m1/c1) × (c2/m2) × λ1 ------------------------(3).

Where c1 and c2 are the order maximum and m = order numbers. Hence;

λ2 = (1/ .350) × (.870/3) × 600 = 497.143 nm.

7 0

3 years ago

Read 2 more answers

f two objects travel through space along two different curves, it's often important to know whether they will collide. (Will a m

Pavel [41]

Answer:

The two objects will collide with the same position vector for all three components at exactly t = 4 s

Explanation:

For two particles starting out at the same time to collide, their position Vector's at the time of collision must be exactly the same.

So, at the collision point, position vector of object 1 is equated to that of object 2.

r₁ = (t², 13t-36, t²)

r₂ = (7t-12, t², 5t-4)

At he point of collision

t² = 7t - 12

t² - 7t + 12 = 0

t² - 4t - 3t + 12 = 0

t(t - 4) - 3(t - 4) = 0

t = 3s or t = 4s

13t - 36 = t²

t² - 13t + 36 = 0

t² - 4t - 9t + 36 = 0

t(t - 4) - 9(t - 4) = 0

t = 9s or 4s

t² = 5t - 4

t² - 5t + 4 = 0

t² - 4t - t + 4 = 0

t(t - 4) - 1(t - 4) = 0

t = 1s or t = 4s

The three components intersect at other times, but at t = 4s, they all intersect at the same time! Meaning that, at this point the two objects are at the same place with the same position vector at that time.

3 0

3 years ago

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directi

DaniilM [7]

Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

4 0

4 years ago

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

IrinaVladis [17]

Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

$\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}$

$\dfrac{P_m1}{P_m2} = \dfrac{2}{1}$

the correct answer is option C

7 0

3 years ago

The reaction mixture represented above is at equilibrium at 298 K. The value of the equilibrium constant for the reaction is 16

andre [41]

Answer : The equilibrium concentration of T(g) is 0.5 M

Solution :

Let us assume that the equilibrium reaction be:

The given equilibrium reaction is,

$R(g)+2T(g)\rightleftharpoons 2X(g)+Z(g)$

The expression of $K_c$ will be,

$K_c=\frac{[Z][X]^2}{[R][T]^2}$

where,

$K_c$ = equilibrium constant = 16

[Z] = concentration of Z at equilibrium = 2.0 M

[R] = concentration of R at equilibrium = 2.0 M

[X] = concentration of X at equilibrium = 2.0 M

[T] = concentration of T at equilibrium = ?

Now put all the given values in the above expression, we get:

$16=\frac{(2.0)\times (2.0)^2}{(2.0)\times [T]^2}$

$[T]=0.5M$

Therefore, the equilibrium concentration of T(g) is 0.5 M

8 0

4 years ago