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WITCHER [35]
2 years ago
5

The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in 1851. Consider a l

ight beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(b) Show that for v<
Physics
1 answer:
Ket [755]2 years ago
8 0

It is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Given ,

The motion of a transparent medium influences the speed of light .

The water moves with speed v in a horizontal pipe .

Assume that the light travels in the same direction as the water moves .

The speed of the light with respect to the water is c/n

Where n = 1.33 is the refractive index of water .

Let us assume ,

u' be the speed of light in water , in the frame moving with the water .

u' is related to the refractive index of water ,n as :

u'=c/n

where , c is the speed of light .

let , u be the speed of light in water in the lab frame .

Now , u and u' are related as : u = (u'+ v )/(1+ u'v/c^2)

Here v is the speed of water in the horizontal pipe .

we know the value of u' , so by substituting the value , we will get ,

u= (c/n+ v)/(1+cv/nc^2)

u= c/n(1+ nv/c)/(1+v/nc)

(b) We have , v<<c

v/c<<1 .

so , (1+v/nc )^-1 = (1-v/nc)

Now substituting this , we will get ,

u = c/n(1+nv/c) (1-v/nc)

u≈c/n(1+ nv/c-v/cn)

u≈c/n + v - v/n^2

Hence , it is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Learn more about speed here :

brainly.com/question/13943409

#SPJ4

Disclaimer : incomplete question , here is the complete question .

Question: The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc) . (b) show that for v<<c , the expression from part (a) becomes , to a good approximation , u ≈ c/n + v - v/n^2 .

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A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
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Answer:

h = 13.3 m

Explanation:

Given:-

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- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

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