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ratelena [41]
3 years ago
15

The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15

cm, what is the distance between the slit and the screen?
Physics
1 answer:
Lady bird [3.3K]3 years ago
3 0

Answer:

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

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How and where is old oceanic crust destroyed?
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8 0
3 years ago
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P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te
tankabanditka [31]

Answer:

(a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

kx-mg=ma

x=\dfrac{ma+mg}{k}....(I)

The spring compressed is \dfrac{ma+mg}{k}.

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

x=2.5\times x_{0}

Here, acceleration is zero

So, x=2.5\times\dfrac{mg}{k}

We need to calculate the acceleration

Put the value of x in equation (I)

2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}

2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)

a=2.5g-g

a=1.5g

Hence, (a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

8 0
2 years ago
Which answer correctly describes the grey lines in this Hering illusion? (Optical Illusions Lesson)
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The answer that correctly describes the grey lines in this Hering illusion is "The grey lines are bent." Option A. This is further explained below

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In conclusion, The statement for the grey lines in this Hering illusion is "The grey lines are twisted.".

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4 0
2 years ago
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A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
chubhunter [2.5K]

Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

<em />

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}

Rmax = 10.001 cm

Therefore, the max volume will be:

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Vmax = 4 190.05 cm^3 = 4.19 L

Now, from (*)

P_{max} = P_1 \frac{V_1T_2}{V_2T_1}

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0
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