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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 646: Lesson Check

Exercise 9

Step 1

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Because the light that is coming form the flashlights consists of many different wavelengths with random phase differences from each other (there is no coherence between the sources) and thus there are no conditions for interference to occur.

Result

2 of 2

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Exercise 10

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They do interfere because they are coherent. Since the path difference between the beams in that particular point is exactly one half of a wavelength they interfere destructively because the destructive interference condition is fulfilled i.e. the path difference is a whole number times half a wavelength.

Result

2 of 2

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Exercise 11

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According to the Huygens’ principle each point of the opening that is being hit by the wave behaves as a source of the wave so when a ray passes through a small opening the waves spread out from its’ every point and their wave front is a sphere around the particular point we count as its’ source.

Result

2 of 2

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Exercise 12

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Superposition of the waves is a rule of calculating the resulting wave in a point in which two or more waves coming from different sources overlap. It says that the current value of oscillating quantity (position of a particle, values of electric and magnetic field, pressure etc.) at a given point is determined simply by adding up all of the values that quantity would have if there have been only the wave coming from one of the sources. Constructive interference happens when the mentioned sum has maximal value, while destructive interference happens when its’ value is minimal.

Result

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Exercise 13

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The spacing would increase. We can easily see that the bigger the angle $theta$ the ray producing a white fringe makes with the central fringe the greater the linear spacing between them. From the condition of diffraction we have

$$
mlambda = dsinthetaRightarrow sintheta = frac{mlambda}{d}.
$$

This means that the sine of the angle $m-th$ ray makes is proportional to $lambda$ and since the sine is an increasing function of an angle (if it falls into the range of $0-90^circ$ which is here the case) so by increasing $lambda$ we increase the angle of deviation of each ray and thus the spacing between them.

Result

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Exercise 14

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The diffraction condition for the 1st bright fringe ($m=1$) reads

$$
lambda=dsinthetaRightarrow sintheta = frac{lambda}{d}Rightarrowtheta=arcsinleft( frac{lambda}{d}right) =0.7^circ.
$$

Result

2 of 2

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Exercise 15

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The diffraction condition for the 2nd bright fringe ($m=2$) reads

$$
2lambda=dsinthetaRightarrow lambda=frac{dsintheta}{2} =frac{48.0times10^{-5}text{ m }timessin(0.0990^circ)}{2} = 415text{ nm}.
$$

Result

2 of 2

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Exercise 16

Step 1

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Vertical distance from central to the first fringe $y$ is related to the distance from the slits to the screen $L$ by

$$
y=Ltantheta
$$
where $theta$ is the interference angle. This yields
$$
tantheta = frac{y}{L}Rightarrow theta=arctanleft(frac{y}{L}right)=arctanleft(frac{0.0536text{ m}}{8.75text{ m}}right) =0.35^circ.
$$

Now writing down the condition for the first bright fringe ($m=1$) we get

$$
lambda = dsinthetaRightarrow d=frac{lambda}{sintheta} = frac{546times10^{-9}text{ m}}{sin35^circ} = 89.4text{ $mu$m}
$$

which is the required separation.

Result

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