Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (u = 0°) is 4.00x10-5 W/m2. What is the intensity at a point on the screen that corresponds to u = 1.20°?

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Hello, fellow physicists today, we're gonna 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. A long narrow aperture of width micrometers is used to to fract monochromatic plane waves of wavelength 514 nanometers. A diffraction pattern is observed on a screen parallel to the slit and 4.00 m from it. The maximum intensity at the bright central fringe is 1.25 multiplied by 10 of the power of negative six watts per meter squared, determine the expected intensity on the screen where the angle of diffraction data is 1.5 degrees. So our angle is to determine the expected intensity on the screen where the angle of diffraction theta is 1.5 degrees. OK. So we're given some multiple choice answers. They're all in the same units of watts per meter squared. So let's read them off to see what our final answer might be. A is 3.71 multiplied by 10 to the power of negative 10 B is 3.05 multiplied by 10 to the power of negative 10 C is 1.69 multiplied by 10 to the power of negative eight. And D is 1.82 multiplied by 10 to the power of negative eight. OK. So first off, we need to recall the equations for the intensity in terms of the phase difference. And we also need to recall and remember the phase difference equation. So let's start off with the equation for intensity. So the equation for intensity states that the intensity is equal to I subscript zero where I subscript zero or I zero is the intensity of the central bright fringe. And that is multiplied by sign of beta divided by two all divided by beta divided by two all squared. OK. And beta represents the phase difference and the phase difference beta is equal to two pi divided by lambda multiplied by a multiplied by sign of theta. OK. Where lambda represents the wavelength and a represents the aperture width. OK. So to begin, let us solver the phase difference first. So let's plug in our known variables. And let's also note that we have to convert nanometers to meters and micrometers to meters. So let's do that. So let's solve for the phase difference. OK. So the phase difference is equal to two pi divided by the wavelength which is given to us as 514 nanometers. But we need to convert that the meter So all we have to do is just take 514 and multiply it by 10 to the power of negative nine to convert it to meters. Then we need to multiply it by the aperture width A which is given to us as micrometers. So we need to convert micrometers to meters. So all we have to do is just take 200 multiply it by 10 to the power of negative six to convert it to meters. And then we need to multiply it by sine of 1.5 degrees because the angle of diffraction in this case, beta is equal to 1.5 degrees. So when we plug that into a calculator, we should get that the phase difference is 64.0 radiance. OK. Now we can solve for the intensity. So let's plug in all of our known values to solve for the intensity. So the intensity is equal to I zero, which the maximum intensity at the central bright fringe in this case is given to us in the problem as 1.25 multiplied by 10 to the power of negative six watts per meter square. And we need to multiply that by sign up. And now we have to take our phase difference value which is 64 radians divided by two. So sign of 64 ratings divided by two all divided by 64 radians divided by two all squared. OK? So when we plug that into a calculator, we should get that our intensity is equal to 3.71 multiplied by to the power of negative 10 watts per meters squared ray. We found our final answer. So when we go and look at our multiple choice answers, that means our correct answer has to be a 3.71 multiplied by to the power of negative 10 watts per meter squared. So that means that the intensity on the screen for a diffraction angle of theta equals 1.5 degrees has to be 3.71 multiplied by 10 to the power of negative 10 watts per meter squared. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Verified Solution

Key Concepts

Diffraction

Intensity of Light

Angular Position in Diffraction Patterns