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zhenek [66]
3 years ago
15

Light of wavelength 578.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 62.5 cm from the slit.

The distance on the screen between the third order minimum and the central maximum is 1.35 cm . What is the width a of the slit in micrometers (μm)?
Physics
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

80.6\mu m

Explanation:

When light passes through a narrow slit, it produces a diffraction pattern on a distant screen, consisting of several bright fringes (constructive interference) alternated with dark fringes (destructive interference).

The formula to calculate the position of the m-th maximum in the diffraction pattern produced in the screen is:

y=\frac{m\lambda D}{d}

where

y is the distance of the m-th maximum from the central maximum (m = 0)

\lambda is the wavelength of light used

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have:

\lambda=578.0 nm = 578\cdot 10^{-9} m is the wavelength

D = 62.5 cm = 0.625 m is the distance of the screen

We know that the distance between the third order maximum (m=3) and the central maximum is 1.35 cm (0.0135 m), which means that

y_3 = 0.0135 m

For

m = 3

Therefore, rearranging the equation for d, we find the width of the slit:

d=\frac{m\lambda D}{y_3}=\frac{(3)(578\cdot 10^{-9})(0.625)}{0.0135}=80.3\cdot 10^{-6} m=80.6\mu m

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8 0
3 years ago
The table below describes some features of methods used to generate electricity. What is method 3?
blondinia [14]

Answer:

dam/hydro power

Explanation:

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2 years ago
Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este
Mariana [72]

a) Density of the liquid: 823.5kg/m^3

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

<em>"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate: </em>

<em>a) the density of the liquid; </em>

<em>b) the weight of the liquid."</em>

a)

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

M=m_L + m_V (1)

where

m_L is the mass of the liquid

m_V is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

m_L=M-m_V=175-35=140 kg

The density of the liquid is given by

d=\frac{m}{V}

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = 0.170 m^3 (volume of the liquid, which is equal to the volume of the vessel)

So we get

d=\frac{140}{0.170}=823.5kg/m^3

b)

The weight of a body is given by

F=mg

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

g=9.8 m/s^2 (acceleration due to gravity)

Therefore, its weight is

F=(140)(9.8)=1372 N

Learn more about density:

brainly.com/question/5055270

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#LearnwithBrainly

6 0
2 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}.

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