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1 year ago

what principle is responsible for alternating light and dark bands when light passes through two or more narrow slits?

1 answer:

7 0

The superposition principle is responsible for alternating light and dark bands when light passes through two or more narrow slits.

The intensity pattern that appears on the lit screen is determined by the superposition principle. When the difference in pathways from the two slits to a location on the screen equals an integral number of wavelengths (0,λ,2λ ,...), constructive interference takes place.

The fact that the two waves' crests follow different paths ensures that they do. A distinctive pattern of brilliant and dark fringes is seen when monochromatic light illuminates a distant screen after passing through two small openings. The superposition of overlapping light waves coming from the two slits results in this interference pattern.

Learn more about superposition principle here;

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2.) Explain why the starting angle doesnt impact the time it takes the pendulum to swing back and forth?

melomori [17]

The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.

The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)

−gsinθ=lθ¨−gain⁡θ=lθ¨

For small angles, θ≪1,θ≪1, and hence sinθ≈θsin⁡θ≈θ. Hence,

θ¨=−glθθ¨=−glθ

This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos⁡(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.

For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause

7 0

2 years ago

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s

ad-work [718]

Answer:

$\theta = 0.0022 rad$

$I = 0.000304 I_o$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

The distance of the slit separation is $d = 0.500 \ mm = 5.0 *10^{-4} \ m$

Generally the condition for two slit interference is

$dsin \theta = m \lambda$

Where m is the order which is given from the question as m = 2

=> $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

substituting values

$\theta = 0.0022 rad$

Now on the second question

The distance of separation of the slit is

$d = 0.300 \ mm = 3.0 *10^{-4} \ m$

The intensity at the the angular position in part "a" is mathematically evaluated as

$I = I_o [\frac{sin \beta}{\beta} ]^2$

Where $\beta$ is mathematically evaluated as

$\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

substituting values

$\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

$\beta = 0.06581$

So the intensity is

$I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

$I = 0.000304 I_o$

3 0

3 years ago

Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m

Radda [10]

Answer:

160J

Explanation:

Given force = 8N and total distance = 20 meters

Workdone = force x distance

= 8 x 20

= 160J

Therefore, workdone by Riley in pulling the hoover is 160J

7 0

3 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$