Light of wavelength 474 nm in air shines on two slits 6.00 x 1...

Question

Light of wavelength 474 nm in air shines on two slits 6.00 x 10^-2 mm apart. The slits are immersed in water, as is a viewing screen 60.0 cm away. How far apart are the fringes on the screen?

Accepted Answer

Hello, fellow physicists today we solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem in a physics lab experiment monochromatic light of wavelength 532 nanometers traveling through air is used to illuminate two slits that are 0.10 millimeters apart. The experiment is conducted in an oil bath with a refractive index of 1.5. And the screen is placed 2.0 m away from the slits, calculate the distance between the consecutive fringes formed on the screen. So that's our angle. Our angle is we're trying to figure out the distance between the consecutive fringes that are formed on the screen. So now that we know that our final answer that we're ultimately trying to solve for our end goal is we're trying to figure out the distance between these consecutive fringes that are formed on the screen. Let's read off her multiple choice answers to see what our final answer might be noting that they're all in the same units of millimeters. So A is 2.8 B is 4.2 C is 5.7 and D is 7.1. OK. So first off, in order to help us solve for this problem, we need to recall and use the formula to help us solve for the angular position theta of the bright fringes from the central fringe which let's call this equation one. And as we should recall, we should be able to write this equation as D multiplied by sine of theta is equal to M multiplied by Lambda prime. Where D is the separation of the slits M is the order of the French and is some integer value. And Lambda prime is the wavelength of light in the oil bath. Awesome. So since we're dealing with an oil bath, we need to note that we're dealing with a setup that isn't oil. So we'll need to use the wavelength in the oil, which let's call this equation to, which means we need to recall that in this particular case that Lambda prime will be equal to lambda divided by lowercase N where lower case N in this case, and let's make a little note of this off the side is the refractive index. And we're told that the refractive index of this oil bath is 1.5 and of course normal Lambda. So we know that Lambda prime, so LAMBDA prime is the wavelength of light in the oil. And lambda by itself is the wavelength of light in the air. Let us also note that we're told that capital L and capital L is the distance of the screen from the slits. And we're told that this value is 2.0 m and this is in the prom itself. And let us also call Y to be the distance of the mth order bright fringe from the central fringe. And as we should recall, a note from small angle approximation, we can go ahead and write that sign of theta is equal to tangent of theta, which is equal to Y divided by L which once again, Y is the distance of the mth order bright fringe from the central fringe. So when we plug this value back into equation one, we will be able to write that D multiplied by Y divided by L is equal to M multiplied by lambda prime, which we can go ahead and expand this and go ahead and write that D multiplied by delta Y divided by L is equal to Lambda prime multiplied by delta M. So let's quickly note here that delta Y is the change in distance and delta M is the change in the order. Awesome. So now moving right along here, so we need to recall a note that for two consecutive bright fringes delta M will have to equal one. So once again, for two consecutive bright fringes delta M will have to equal one. So let's call this equation three. So anything multiplied by one will be itself. So we could go ahead and write that delta Y is equal to Lambda prime multiplied by L all divided by D. Fantastic. So next order of business is we need to use equation three. Sorry, we need to take equation two and we need to plug it into equation three. And when we do that, we will be able to write that delta Y is equal to lambda multiplied by L divided by N multiplied by D. So now all we need to do is we need to plug in all of our known variables into our delta Y equals lambda multiplied by L all divided by N multiplied by D. And when we do that, we will find our final answer for delta Y which is the final answer that we're ultimately trying to solve for. So plugging in all of our known variables. So we know that the wavelength is equal to 532 nanometers. But let's convert nanometers to meters. So we need to take 532 and we need to multiply it by 10 to the power of negative nine Fantastic. And then we need to multiply it by our value of L which is 2.0 m. Then we need to divide everything by the refractive index which is 1.5. And we need to multiply it by our value of D. And we're told that D is equal to 0.10 millimeters. But we need to convert millimeters to meters. So we need to take 0.10 mean to multiply it by 10 to the power of negative 3 m. So I've gone ahead and done all the conversions for you and in an effort to save time solving, but you can easily do this conversion on your own using dimensional analysis. But in an effort to save time, I've just given you the conversions. So when we plug this into our calculator, we will find that our final answer for delta Y will be equal to 7.09 multiplied by 10 to the power of negative 3 m. And that's when we round to two decimal places. But we have a bit of a problem. Our final answer is in units of meters, but we need to convert meters to millimeters. So once again, you can use dimensional analysis. But when we, when we, when we convert to units of millimeters, and we round to two significant figures or to one decimal place to match all of our multiple choice answers, we will find that delta Y is approximately equal to 7.1 millimeters. And that is our final answer. Hooray, we did it. So that's it. So looking at our multiple choice answers, the correct answer has to be the letter D which once again is 7.1 millimeters. So this is a very easy problem to solve. For as long as you remember how to properly convert your units and understand the conceptual physics behind this type of problem. So, you know, you're so you're able to like quickly know and recall all the equivalent equations that are all the proper equations that you need to use to solve for a problem like this. So all the relevant, there's the, there's the word else with all the relevant equations that you need to use to solve for this particular problem and that's it and that's all. So thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Light of wavelength 474 nm in air shines on two slits 6.00 x 10^-2 mm apart. The slits are immersed in water, as is a viewing screen 60.0 cm away. How far apart are the fringes on the screen?

Key Concepts

Interference of Light

Young's Double Slit Experiment

Refraction and Wavelength in Different Media

Watch next

Light of wavelength 474 nm in air shines on two slits 6.00 x 10-2 mm apart. The slits are immersed in water, as is a viewing screen 60.0 cm away. How far apart are the fringes on the screen?

Key Concepts

Interference of Light

Young's Double Slit Experiment

Refraction and Wavelength in Different Media

Watch next

Light of wavelength 474 nm in air shines on two slits 6.00 x 10^-2 mm apart. The slits are immersed in water, as is a viewing screen 60.0 cm away. How far apart are the fringes on the screen?