All categories

2 years ago

Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.

Physics

1 answer:

KIM [24]2 years ago

6 0

Answer: the airy pattern can only arise from wave propagation

Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs

You might be interested in

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy

AURORKA [14]

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

8 0

3 years ago

A professor drives off with his car (mass 870 kg), but forgot to take his coffee mug (mass 0.47 kg) off the roof. The coefficien

SSSSS [86.1K]

Answer: $6.86 m/s^2$

Explanation:

Given

mas of car=870 kg

coffee mug mass=0.47 kg

coefficient of static friction between mug and roof $\mu _s= 0.7$

Coefficient of kinetic Friction $\mu _k=0.5$

maximum car acceleration is $\mu \times g$

here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

$a_{max}=0.7\times 9.8=6.86 m/s^2$

8 0

3 years ago

In a series circuit, when batteries are added to the circuit the voltage will _____ and the current will _____.

Eva8 [605]

Answer: voltage, explode is the answers

Explanation:

Bc of the device for each server

5 0

3 years ago

You lift a 50 N object 2 meters off the ground what work did you do on the object

mixer [17]

Work equals force × displacement (distance between initial point and end point is displacement)
if u follow this it becomes
work = 50 × 2 which is equal to 100

comment if u have more questions

4 0

3 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$