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Natalka [10]
3 years ago
15

laser light sent through a double slit produces an interference pattern on a screen 3.0 m from the slits. If the eighth order ma

ximum occurs at an angle of 12.0, at what angle does the third order maximum occur
Physics
1 answer:
Xelga [282]3 years ago
8 0

Solution :

Given the laser light which is sent through the double slit produces an interference pattern on the screen placed 3  meters from the slits.

The 8th order maximum occurs at angle = 12

So,

$8^{th} \text{ order maxima} = d \sin \theta = m \lambda$      , m = 8

$d = \frac{8 \lambda}{\sin 12}$

$\frac{\lambda}{d}= \frac{\sin 12}{8}$

$3^{rd} \text{ order maxima}= d \sin \theta_2 = m_2 \lambda$

$\sin \theta_2 = \frac{m_2 \lambda}{d}=\frac{3 \lambda}{d}$ $=0.75\ {\sin 12}$

$\theta_2 = \sin^{-1}\left(0.75\ \sin 12\right)$

   $ = \sin^{-1}\left(0.155)$

   $=8.91^\circ$

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How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.
goldfiish [28.3K]
143m/s if you just perhaps by what you know you'll figure it out
4 0
3 years ago
When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc
Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

          w² = mg d / I

In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

         d = L / 2

The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

        I = ¼ m r2 + ⅓ m L2

        I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

         V = π r² L

          m =  ρ π r² L

let's substitute

        w² = m g (L / 2) / m (¼ r² + ⅓ L²)

        w² = g L / (½ r² + 2/3 L²)

        L >> r

         w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0
3 years ago
A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
balandron [24]

Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

Hence:

\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

In other words, the angle of refraction in the glass would be approximately 28^{\circ}.

7 0
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