The wavelength range of the visible spectrum is approximately ...

Question

The wavelength range of the visible spectrum is approximately 380–750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in the first order.

Accepted Answer

Hello, fellow physicists today, we're going to solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. Visible light is a form of electromagnetic wave. Each color is it is of a different wavelength red has the longest wavelength lambda equals 720 nanometers and violet has the shortest one lambda equals nanometers. If ordinary white light from a source is incident normally on a slits per millimeter to fraction grating. What should be the angular spread for the second order of the visible spectrum? Awesome. So we're given some multiple choice answers here and they're all for the angular spread which is denoted as theta subscript S and they're all in the same units of degrees. Let's read them off to see what our final answer might be. A is 7.8 B is 10.4 C is 8.4 and D is 9.2. OK. So first off, let us recall the equation to determine the maximum intensity. And let's call it equation one So the equation defined the maximum intensity is lower case D which is the width length of the slit slit is equal to sine, theta is equal to N which is some integer value for the order number multiplied by lambda. So the maximum intensity is the width length of the slit is equal to sine theta which is equal to the nth order multiplied by the wavelength represented by LAMBDA. OK. So to solve for the slit width, we need, we can use one divided by D. So let me do that one divided by D equals 220 slits per millimeter. And now we need to rearrange this equation to solve for D. So when we rearrange this equation to solve for D, we should get that D equals one millimeter divided by 220. So when we plug that into a calculator, we should get 4. multiplied by 10 to the power of negative three millimeters. But we need to convert this value to meters. And to do that, all we have to do is take 4.55 and multiply it by 10 to the power of negative six to get it in meters. Awesome. So now we can rewrite equation one as, and we'll call it equation two as sine theta equals the nth order multiplied by the wave length divided by the width length of the slit. OK. So now we need to use equation two to solve for the violet and red wavelengths. So note that we can rewrite equation to, to solve VERS theta by itself. And to do that, all we have to do is take sign in verse multiplied by the nth order multiplied by the weight length divided by the distance or I should say the width length of the slip. OK. So now we need to plug in our known values to solve for the theta value for violet light, I should say or the violet wavelength condition. So we also call it theta subscript V. So and it's for the second order. So N in this case will equal two, I'll write in a quick little note here, N equals two. So two multiplied by the wavelength which the wavelength was given to us for violet light as 400. But we need to convert nanometers to meters. So we need to take 400 multiplied by 10 to the power of negative 9 m divided by the width length of the slit which is given to us. Oh, actually, it's the value we just found as 4.55 multiplied by 10 to the power of negative 6 m. OK. So when you plug that into a calculator, we should get that data subscript V equals 10.1 degrees. OK. So we need to do the same thing defined our beta subscript R for the red light wave length that I that we were given. And the problem that we need to solve for. So beta subscript R so for the red light wavelength, it's sign inverse. So this is plugging in all our known variables. And so we're also doing in this case, it's N equals two means it second order. So two multiplied by the wavelength of red light, which is given to us as 720 nanometers. But like we did before with the convert 2 m, so 720 multiplied by 10 to the power of negative 9 m divided by our devalue, which we found which was 4.55 multiplied by 10 to the power of native 6 m. So when we plug it into a calculator, we should get that theta subscript R is equal to 18.5 degrees. OK. So do you determine the angular spread which we're gonna call it theta subscript S? So the angular spread we must, all we must do is subtract theta subscript R and theta subscript V like. So, so the angular spread is equal to theta subscript R minus theta subscript V. So let's plug in our known values to solve for the angular spread. So 18.5 degrees minus 10.1 degrees equals 8.4 degrees. So our angular spread equals 8.4 degrees. Awesome. And that is our final answer. We did it. So that means our final answer has to be C 8.4 degrees. So the angular spread is equal. So theta subscript S is equal to 8.4 degrees. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

The wavelength range of the visible spectrum is approximately 380–750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in the first order.

Key Concepts

Diffraction Grating

Order of Diffraction

Angular Width of the Spectrum

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