All categories

3 years ago

Which best describes a difference between laser light and regular light?

1 answer:

4 0

The basic difference is that the ordinary sources are incoherent that means that the discrete frequencies merge up to give an intermediate between the maximum and minimum frequencies. While the laser is coherent containing the single frequency with maximum amplitude. thus travelling far.

You might be interested in

Light propagate faster through medium “a” than medium “b”

dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1, n_2$ are the index of refraction of the two mediums

$\theta_1, \theta_2$ are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

$sin \theta_1 = sin \theta_2$ (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

$n_1\neq n_2$

Therefore the only angle that can satisfy eq.(1) is

$\theta_1 = \theta_2 = 0$

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

3 0

3 years ago

When is a white dwarf formed?

Anna [14]

When a red giant has insufficient mass it pretty much sheds its outer layer to leave the center core know as the white dwarf

6 0

3 years ago

Read 2 more answers

Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine

TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because, the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by $tanθ = y/D$

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ = λD/d and y₋₁ = -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ = 3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ = λD/d and y₂ = 2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0

3 years ago

Okay so I've have 10 Brainliest, through my whole time using brainly- I want people to get brainly and points so Im giving away

Nookie1986 [14]

answer:

thank youu for the points cute drawing by the way (✿◠‿◠)

when did you join brainly though?

have a wonderful dayyy !!

4 0

3 years ago

Read 2 more answers

Question 9 of 27

Ksju [112]

The power of man performing 500 J of work in 8 seconds is 62.5 J/s.

Power can be defined as the pace at which work is completed in a given amount of time.

Horsepower is sometimes used to describe the power of motor vehicles and other machinery.

The pace at which work is done on an item is defined as its power. Power is a temporal quantity.

Which is connected to how quickly a project is completed.

The power formula is shown below.

Power = Energy / Time

Power = E / T

Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.

Here we have given Energy as 500 J and Time as 8 second.

Power = Energy / Time

Power = 500 / 8 Joule / sec

Power = 250 / 4 Joule / sec

Power = 125 / 2 Joule / sec

Power = 62.5 Joule / sec or 62.5 watt

Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.

So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.

Learn more about Power here:

brainly.com/question/1634438

#SPJ10

6 0

2 years ago