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pickupchik [31]
3 years ago
15

When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is

travelling from media 1 to media 2.
At what angle of incidence is the reflected ray perpendicular to the incident ray? The indexes of refraction for the two media are n1 and n2, respectively.

Express your answer in terms of the variables n1 and n2
Physics
1 answer:
podryga [215]3 years ago
3 0

Answer: critical angle, sin^-1 (n2/n1)

Explanation: the angle of incidence at which the retracted ray makes an angle of 90° with the normal is known as the critical angle.

Snell's law defined refraction mathematically as shown below

n1 sin θi = n2 sin θr

n1 = refractive index of the first medium

n2 = refractive index of the second medium

θi = angle of incidence

θr = angle of refraction

When the refrafted ray is perpendicular to the normal, the angle of refraction (θr) is 90° hence making the angle of incidence (θi) the critical angle θc

By substituting these conditions into the Snell's law, we have that

n1 sin θc = n2 sin 90

According to trigonometry, the value of sin 90 is 1, hence we have that

n1 sin θc =n2

sin θc = n2/n1

θc = sin^-1 (n2/n1)

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The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

<u>Explanation: </u>

Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.  

           \text { Work done }=\text { Force } \times \text { displacement }

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

           \text { Work done }=200 \mathrm{N} \times 0=0

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.

8 0
3 years ago
Which of the following energy resources produces greenhouse gases?
Blizzard [7]
Coal is burned to get energy. By burning coal, green houses gases like carbon dioxide, nitrogen oxides and suphur oxides are produced.

Rest of the energy sources given in the option are clean energies.

Answer is A.
4 0
3 years ago
On March 21, a stick casts the following shadow. What is the most likely time of day?
jenyasd209 [6]

Answer:

9:00 AM

Explanation:

I took the test and that was the answer

5 0
3 years ago
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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.
KIM [24]

Answer: the airy pattern can only arise from wave propagation

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6 0
2 years ago
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