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3 years ago

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When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin

gle rectangular region of the viewing screen is illuminated (matching the shape of the thin rectangular slit).
O True O False

1 answer:

saveliy_v [14]3 years ago

7 0

True
If it helps you plz brainlest me

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Two north poles will blank

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Be pushed away from each other.

Explanation:

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Which image illustrates the bouncing of a light wave off of a surface?

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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

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3 years ago

1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com

Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

$where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s$

$d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m$

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

$d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}$

$d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}$

$d' = 594\sqrt{1 - 0.81}$

$d' = 594 \times 0.4359$

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0

3 years ago