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s2008m [1.1K]
3 years ago
10

What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum fo

r visible light? Assume visible light has a range of 400 nm to 700 nm.
Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

Answer:

14,300 lines per cm

Explanation:

Answer:

14,300 cm per line

Explanation:

λ400 nm to 400nm

We can find the maximum number of lines per centimeter, which is reciprocal of the least distance separating two adjacent slits, using the following equation.

mλ = dsin (θ)

In this equation,

m is the order of diffraction.

λ is the wavelength of the incident light.

d is the distance separating the centers of the two slits.

θ is the angle at which the mth order would diffract.

To find the least separation that allows the observation of one complete order of spectrum of the visible region, we use the maximum wavelength of the visible region is 700 nm.

d =  mλ / sin (θ)

As we want the distance d to be the smallest then sin (θ) must be the greatest, and the greatest value of the sin (θ) is 1. For that we also use the longest wavelength because using the smallest wavelength, the longest wavelength would not be diffracted.

d =  mλ / sin (θ)

d =  1 x 700nm / 1

  = 700 nm

So, the least separation that would allow for the possibility of observing complete first order of the visible region spectra is 700 nm, and knowing the least separation we can find the maximum number of lines per cm, which is the reciprocal of the number of lines per cm.

n = 1/d

   = 1 / 700 x 10^{-9}

  = 1, 430,000 lines per m  

  =  14,300 lines per cm

<u>The maximum number of lines per cm, that would allow for the observation of the complete first order visible spectra.</u>

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Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

8 0
3 years ago
A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func
OlgaM077 [116]

Answer:

A= 148.92  m/s²

Explanation:

Given that

U(x,y) = (6.00  )x²  - (3.75  )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

A=\sqrt{a^2+a'^2}

A=\sqrt{120^2+88.2^2}

A= 148.92  m/s²

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3 years ago
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LekaFEV [45]

Answer:

It will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

Explanation:

From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.

To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.

This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.

Now,

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x = (160 inches × 1 day) / 36 inches

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Hence, it will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

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