Question Video: Finding the Distance Between Two Slits Using Other Variables | Nagwa Question Video: Finding the Distance Between Two Slits Using Other Variables | Nagwa

# Question Video: Finding the Distance Between Two Slits Using Other Variables Physics • Second Year of Secondary School

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Light with a wavelength of 636 nm passes through a sheet in which there are two parallel normal slits. The light from the slits is incident on a screen parallel to the sheets, 1.08 m away, where a pattern of light and dark fringes is observed. A line πΏ runs perpendicular to the surface of the sheet and the direction of the slits. The line πΏ intersects the central bright fringe of the pattern on the screen. The distance on the screen from πΏ to the center of the bright fringe nearest to the central bright fringe is 6.11 cm. What is the distance between the slits? Give your answer to two decimal places in scientific notation.

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### Video Transcript

Light with a wavelength of 636 nanometers passes through a sheet in which there are two parallel normal slits. The light from the slits is incident on a screen parallel to the sheets, 1.08 meters away, where a pattern of light and dark fringes is observed. A line πΏ runs perpendicular to the surface of the sheet and the direction of the slits. The line πΏ intersects the central bright fringe of the pattern on the screen. The distance on the screen from πΏ to the center of the bright fringe nearest to the central bright fringe is 6.11 centimeters. What is the distance between the slits? Give your answer to two decimal places in scientific notation.

This problem is essentially trigonometry once we set up a diagram describing it, but a large part of the difficulty is setting up this diagram correctly in the first place. Letβs take what the question tells us step by step. Light with a wavelength of 636 nanometers passes through a sheet in which there are two parallel narrow slits. The light from the slits is incident on a screen parallel to the sheets, 1.08 meters away, where a pattern of light and dark fringes is observed. A line πΏ runs perpendicular to the surface of the sheet and the direction of the slits. The line πΏ intersects the central bright fringe of the pattern on the screen.

Now then, this next sentence here is where things can get really tricky. The distance on the screen from πΏ to the center of the bright fringe nearest to the central bright fringe is 6.11 centimeters. To see exactly what this means, letβs zoom in on the area where πΏ intersects the central bright fringe. The central bright fringe, which is this one here, and also just zoomed in this one right here is the fringe directly opposite the center of the two slits over here.

Because line πΏ runs exactly perpendicular to the center of these slits, it means that the line πΏ intersects the central bright fringe, but also specifically the center of the central bright fringe. This is important to know because when we want to measure the distances between the bright fringes, we want to measure from the centers of those bright fringes, since these are the most distinct and easy to measure areas.

So when this sentence tells us that the distance on the screen from πΏ to the center of the bright fringe nearest to the central bright fringe is 6.11 centimeters, it means weβre looking at the distance between the center of the central bright fringe and the center of the other bright fringe nearest to it, though the one nearest to the central bright fringe is actually either the one directly above or below it. They are equidistant. And our calculations will remain the same regardless of which one we choose. So, weβre going to arbitrarily choose the top bright fringe.

The question text tells us that the distance between the center of these fringes is 6.11 centimeters. Now then, weβve set up our diagram using what the question has told us, and now we just have to find the distance between these slits. Weβre now going to draw another zoomed-in diagram that shows the distance between these slits. So weβre going to erase some of the question text. Now then, we have the screen with the two slits in it, and weβll call the distance between the slits π.

When this light wave passes through the slits, it splits, forming the interference pattern on the screen opposite the screen with the slits. What we want to do is find out which of these light waves produce the bright fringe nearest to the central bright fringe, which is the one that is 6.11 centimeters away from the central bright fringe. We represent this by having two light waves coming from the slits at an angle. If we form some lines parallel to the screen with the slits in it, then we can find this angle, which weβll call π for both the top and bottom light waves from the slits.

Now, this is not strictly true, since when going to the first bright fringe away from the central bright fringe, these light waves do eventually converge, meaning that the angles have to be very slightly different. But the difference is so minuscule it is safe to say that they are basically the same angle of π. Now then, weβre going to do something clever using our knowledge of the areas where constructive interference occurs in an interference pattern.

Constructive interference occurs when the path length difference between two waves of the same wavelength π is ππ, where π is an integer. The path length difference is exactly that, the difference in length that the paths of the two waves take. For the case of the central bright fringe, the path length difference is actually zero, since both waves are traveling the same distance from the slits, which means that the central bright fringe is where π is equal to zero, meaning that there is no path length difference, since of course zero times π is just zero.

But letβs now look at the case where we look at the bright fringe nearest to the central bright fringe. Since this is the next closest bright fringe, π is equal to one, meaning that the path length difference is one times π, or just π.

Now that we have this information, we can go back to looking at the diagram up here. If we assume that these two light waves here are going towards the first bright fringe away from the central bright fringe, which is again this one here, then we know that the path length difference between these two light waves is equal to π. And we know that this path length differences applied to the lower wave since it has to travel a further distance. Knowing this is very handy because now we can actually form a right triangle with π and π by drawing a line straight down from the top of π, the top slit, down to the path length difference, forming a right angle.

So, we know the value of π, which is 6.36 nanometers. But to find π using this triangle and subsequent trigonometry, we would have to find an angle. And luckily, we already know what this angle is here. This angle is actually π. So, if we find the value of π, we can use it combined it with the wavelength to find the value of π.

And to find π, we can do another clever trick involving this line πΏ. πΏ is the line going straight perpendicular from the center of the slits, which would look like this on the diagram above. Since both of these angles here are similar enough to be called the same, of just π, then if we draw a light wave from the center of the slits to the first bright fringe from the central bright fringe, it too will have that same angle π. And we can determine this angle because we know the length πΏ, which is just 1.08 meters, the distance between the screens, and we know the distance between the center of the central bright fringe and the first bright fringe away. This all together forms a right triangle, allowing us to use the relation of tangent π is equal to the opposite side over the adjacent side.

The opposite side is 6.11 centimeters, and the adjacent side, πΏ, is 1.08 meters. To make things convenient for ourselves, weβre going to convert these centimeters to meters by dividing them by one meter over 100 centimeters. The centimeters cancel, leaving us with 6.11 times 10 to the power of negative two meters. We can then substitute in these numbers and divide 6.11 times 10 to the negative two meters by 1.08 meters, leaving us with about 5.657 times 10 to the negative two meters.

We can then find the angle π by taking the inverse tangent of both sides, which, by using our calculator, we can find out that the inverse tangent of 5.657 times 10 to the negative two meters is about 3.238 degrees. Of course, this number keeps going, so donβt round yet until you get your final answer.

Now that we know π and π, we can find π by using a sine relation. Since for a right triangle, sin π is equal to the length of the opposite side of the angle π over the length of the hypotenuse. The opposite side from the angle π is π, and the hypotenuse is π. Substituting in these values gives sin π is equal to π over π. We want to find π by itself. So, letβs multiply both sides by π in order to cancel it in the denominator on the right side. This gives us π sin π is equal to π. So now to get π by itself, letβs divide both sides by sin π. This causes the sin πβs on the left side to cancel, leaving behind π is equal to π over sin π. π is 6.36 nanometers. And we already found π as 3.238 degrees.

Because we want the answer in scientific notation, it would be wise to put the value of π in scientific notation as well. Nanometers are 10 to the power of negative nine meters, meaning that it is 636 times 10 to the power of negative nine meters, or 6.36 times 10 to the power of negative seven meters in scientific notation. Using our calculators, this number over the sin of 3.238 degrees in scientific notation to two places is equal to 1.12 times 10 to the power of negative five meters. This is the distance between the slits.

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