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Ymorist
2 months ago
5

Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizon

tally, and close enough to each other so that both of them fall in the beam. A precise positioning shield is used to cover alternatively one of the slits, while the other remains open. On an observation screen positioned 1 meter further from the opaque screen, the diffraction patterns from the independent slit illumination were found identical, with minima 4.0 mm apart. The interference pattern maxima from simultaneous illumination of both slits (shield removed) were 1.0 mm apart. What are the double slit characteristics (slit width, slit separation)? Provide the result in millimeters.
i found that the answers for part 1: slit width= 0.158 mm slit separation= 0.633 mm

problem 2: On the setup described in problem 1, the opaque screen with slits was replaced with a diffraction grating with 20 slits/mm. What distance between the diffraction maxima is to be expected on the observation screen?
Physics
1 answer:
Softa [3K]2 months ago
6 0

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

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inna [3103]
1. f1 = Vs + Vbat Vs fO
2. f1 = Vs Vs - Vbat fO
3. f1 = Vs Vs + Vbat fO
4. f1 = Vs + Vbat Vs - Vbat fO
5. f1 = 2Vs Vs - Vbat fO
6. f1 = Vs - Vbat Vs fO
7. f1 = Vs - Vbat Vs + Vbat fO
8. f1 = fO
9. f1 = 2Vs Vs + Vbat fO

The frequency that is detected is
f= (V+Vr) × fO/ (V + Vs)
The speed of sound is V = Vs
The speed of the receiver is Vr = O
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4 0
2 months ago
A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?
inna [3103]

Answer: 24.24 m

Explanation:

A player launches a football 50.0 m at an angle of 61° to the north of west. We will break this down into vertical and horizontal elements.

Horizontal component: 50 cos 61° = 24.24 m directed westward

Vertical component: 50 sin 61° = 43.73 m directed toward the north.

Refer to the diagram below.

Therefore, the westward displacement of the football corresponds to the horizontal component of the displacement, which is 24.24 m.

7 0
1 month ago
A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc
inna [3103]

Answer:

17 m/s south

Explanation:

m_1 Mass of the dog = 10 kg

m_2 Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

u_1 Velocity of the dog = 1 m/s

u_2 Velocity of skateboard

In this scenario, linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

The speed of the skateboard post-dog jump will be 17 m/s south, considering north as the positive direction

3 0
1 month ago
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Sav [3153]
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
1 month ago
Jo, Daniel and Helen are pulling a metal ring. Jo pulls with a force of 100N in one direction and Daniel with a force of 140N in
inna [3103]

Answer:

She exerts a force of 40 N.

Explanation:

The fact that the ring remains stationary indicates that the forces are in equilibrium.

Let’s denote Jo's force as x.

The equation to consider is

140 = x + 100

x = 40

5 0
1 month ago
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