If you double the width of a single slit, the intensity of the...

Question

If you double the width of a single slit, the intensity of the light passing through the slit is doubled. (a) Show, however, that the intensity at the center of the screen increases by a factor of 4. (b) Explain why this does not violate conservation of energy.

Accepted Answer

Hello, fellow physicists today, we're gonna 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, light passes through a rectangular aperture of a single slit in an opaque screen, creating a diffraction pattern. On a distant screen, the slit initially has a certain width and the intensity of the light at the central maximum is measured. I if the width of the slit is halved, determine the intensity at the central maximum I I determine the total energy in the diffraction pattern when the width of the slit is halved. Awesome. So it appears for this particular problem we're asked to solve for two separate answers. For part I, our first answer, we're asked to figure out the intensity at the central maximum when the width of the slit is halved. And for our second answer, for part I I, we're asked to determine the total energy in the diffraction pattern when the width of the slit is halved. So now that we know that we're solving for two separate answers, let's read off multiple choice answers to see what our final answer pair might be noting for part I, it will state that I knew equals sum expression. And for part I, I will state that E total is equal to sum expression and I'll read them in order for each multiple choice answer. So for a, it is I zero divided by four and E zero divided by two, B is I zero, divided by two and E zero, C is I zero divided by four and E zero. And finally, D is I zero divided by two and E zero, divided by two. OK. So first off, we need to recall and understand diffraction patterns. So we need to have an understanding of the conceptual physics that's behind diffraction patterns. So as we should recall, when light passes through a single slit, it creates a diffraction pattern on a screen, the central maximum is the brightest part of this pattern and the intensity decreases as you move away from the center. We also should recall that the intensity of the light at the central maximum is determined by the electric field amplitude produced by the light that is passing through the slit. Awesome. So now at this point, we need to start recalling the effect that having a slit width will do. So we need to recall what effect when you take the width of the slit and you have it. So what effect will that have if you make the slit and you have it when you take the slit and you have it, meaning you cut the slit in half. What is, what happens when you cut the slit in half? So as we should recall, the slit is initially at a certain width and now we are having that width, meaning we're cutting that width in half, we're taking half the width. So as we should recall, the electric field amplitude at the central maximum is the sum of contributions of or from I should say from the wavelets of light that are passing through the slit. So once again, this is very important. As we should recall, the electric field amplitude at the central maximum is the sum of the contributions from these wavelets of light that are passing through the slit. So when the slit is ha or the slit width is halved, fewer wavelengths wavelengths, these wavelengths of light will contribute to the total electric field amplitude at the center. So this is very important. So when we cut our slit in half fewer wavelength wavelets, which wavelets is just a fancy way of saying wavelengths will contribute to the total electric field amplitude at the center. Awesome. So now moving right along here, our next step is we need to recall and use the relationship between the electric field and the intensity. So we need to recall a note that the intensity I of light at any point will be proportional to the square of the electric field amplitude A which putting what we just said into w what we just said in words into writing will say that the intensity I is proportional to using our proportional symbol is proportional to A squared where A squared once again is our electric field amplitude that has been squared. So if the slit width is halved, the total electric field amplitude at the center also decreases as we should recall because fewer wavelengths wavelengths, these wavelengths will contribute to the central maximum. Awesome. So now our next step is we need to calculate the change in intensity. So as we should recall once again, so when the slit width is halved, the electric field amplitude at the center will also be halved. Meaning that the new electric field amplitude will become a new is equal to a zero divided by two where a zero is the initial electric field amplitude. So therefore, we can go ahead and write that the intensity at the central maximum will now be proportional to I knew. So I knew our new intensity will be proportional to using our proportional symbol again, will be proportional to a zero divided by two. So a zero divided by two all to the power of two, which will be equal to a zero squared divided by four, which will be all equal to I zero, divided by four. And that is our final answer. For part I hooray, we did it. Now we could start solving for part I I, our second answer. Awesome. So first off, in order to solve for part I I, our second answer, we need to define the initial energy and the slit width. So let us assume that the initial width of the slit is a zero. And this time, we need to make sure that we're specifying that we're dealing with lowercase A zero and A zero. In this case, lowercase A zero is the initial width of the slit. And the total energy passing through the slit I initially is E zero. Fantastic. So the relationship between the slit width and the energy passing through the slit can be expressed as, as we should be able to recall that E zero is proportional using our proportional symbol once again to lowercase A zero. Or we can go ahead and write that E zero is equal to K specifically lowercase K multiplied by lowercase A zero where K is some constant value. So now our next step is we need to have the slit width. So we need to take half of the slit width. So the problem tells us as we should recall and note that the slit width is halved. So therefore, we could go ahead and write that lowercase A new is equal to a zero, squared words lowercase A zero square. So a zero divided by two, I said squared by I meant to say divided by two. So a new lowercase a new is equal to lowercase A zero divided by two. Thus, we can go ahead and write that E new is equal to K multiplied by a lowercase A new which is equal to K multiplied by A zero, divided by two, which is all equal to E zero, divided by two. And that is it we've officially solved for this problem. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter A which once again for I is I new equals I zero, divided by four and IIE total equals E zero divided by two. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

If you double the width of a single slit, the intensity of the light passing through the slit is doubled. (a) Show, however, that the intensity at the center of the screen increases by a factor of 4. (b) Explain why this does not violate conservation of energy.

Key Concepts

Single Slit Diffraction

Intensity of Light

Conservation of Energy

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