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laiz [17]
2 years ago
8

a film of transparent material 120 nm thick and having refractive index 1.25 is placed on a glass sheet having refractive index

1.50. (A) Determine the longest wavelength of light that interferes destructively when reflected from the film. (B) Determine the longest wavelength that interferes constructively
Physics
1 answer:
Eva8 [605]2 years ago
4 0

Answer:

a) 600nm

b) 300nm

Explanation:

the path difference = 2t  

t = thickness of the film

L' = wavelength of light in film = L/n

L = wavength of light in air

n = refractive index of glass

(a)

for destructive interference 2t = L'/2 = L/2n

L = 4*t*n

= 4*120*10^-9*1.25  

L = 600 nm

(b)

for constructive interference 2t = L' = L/1.25

L = 2tn

= 2 × 1.25 ×  120nm

= 300 nm

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A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

80=5(50-x)

80=250-5x

5x=170

x=\frac{170}{5}

x=34 cm

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2 years ago
How many top quark lifetimes have there been in the history of the universe (i.e., what is the age of the universe divided by th
ololo11 [35]

Answer:

1.0\cdot 10^{41} times

Explanation:

First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.

Age of the universe:

T=100,000,000,000,000,000s = 1.0\cdot 10^{17} s (1 followed by 17 zeroes)

Lifetime of the top quark:

\tau = 0.000000000000000000000001s = 1.0\cdot 10^{-24} s (we moved the decimal point 24 places to the right)

Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:

r = \frac{T}{\tau}=\frac{1.0\cdot 10^{17} s}{1.0\cdot 10^{-24} s}=1.0\cdot 10^{41}

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2 years ago
To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes a
Gwar [14]

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

5 0
2 years ago
The definition of theory says that a it is a hypothesis or group hypotheses. In your own words, explain why a theory is also a h
BigorU [14]
A hypothesis is a tentative and testable explanation, based on observation(s). A hypothesis can be supported or refuted through experimentation or more observation. A hypothesis can be disproven, but not proven to be true.
8 0
2 years ago
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A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy
skad [1K]

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

\theta= \omega t

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\omega is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of \omega is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

(b) The girl has greater linear speed.

Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the point from the centre of the merry-go-round

In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of \omega is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.

3 0
2 years ago
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