Pset 1 Solutions PDF

6.2540 PSET 1 Solution Q2 Single Slit Diffraction Part a) From lecture, we have that the minima for a single slit diffraction pattern occurs at position on the m th...

6.2540 PSET 1 Solution Q2 Single Slit Diffraction Part a) From lecture, we have that the minima for a single slit diffraction pattern occurs at position on the m th x screen given the following formula: mλL x = (1) D Rearranging for the slit width, we get: mλL D = (2) x Since we are looking at the position of the first minima, we have that 1, m = x = 1.75 × 10 −2 m , λ = 580 × 10 −9 and m L = 8m. Plugging in the numbers to the equation above, we get that D ≈ 265.1μm. Part b) From the above equations, we see that is inversely proportional to. Therefore, if the slit width D x decreases, the spacing between minima will get wider. Part c) From the above equations, we see that is proportional to. Therefore, as increases, so does the L x L spacing between minima. Part d) Using the equation from part (a) for the slit width, we can evaluate the maximum and minimum slit widths to be Dmin ≈ 264.4 μm and Dmax ≈ 265.9 μm. Therefore, D = 265.1±0.75 μm. Q3 Lenses and imaging Part 1 Part a) Rearranging the thin lens equation, we get: 1 d0 − f = (3) di f d0 f d0 ∴ di = d0 − f If we set d0 → f then: di → ∞ (4) This means that an image would only be formed at infinity, that is, no image will be formed. Part b) The linear magnification of a lens is simply the ratio of the height of the formed image and the original image, i.e. M =. We can figure out this ratio by tracing the rays emanating from the top of the object hi that cross through the middle of the lens (which will be undisturbed) and a ray perpendicular to the lens ho (will pass through the focal point) as in the diagram below: We can see that the triangle formed by the height of the object, the optical axis (middle line perpendicular to the lens) and the ray from the top of the object that passes through the middle of the lens will be a similar triangle to the one formed by the height of the image and the same two other rays. We can confirm this by the fact that the angle between their hypotenuse and the optical axis is exactly the same. This means that the ratio of any two similar sides of the triangle must be the same, therefore: hi di M = = − (5) ho do where we have used a negative sign to indicate the convention that the image formed will be inverted. From part a) we then get that: di f M = − = (6) do f − do Part c) Rearranging for in the previous equation we get: do f (M − 1) do = (7) M Given that the magnification can be positive or negative, for we have that |M | = 10 or do = 90. do = 110 Similarly, for |M | = 100 we have that or do = 99. We can see that, as increases, do = 101 |M | do approaches the focal point. That means that, to get high magnifications, we would need the object to get asymptotically closer to , which could make alignment in real life very tricky and sensitive or do = f practically impossible. Part d) From the previous equation, we can see that, if changes, so will. But, from the thin lens equation in M do part a), we know that, whenever changes, so will. Since we want to place the detector at so that the do di di image is formed at the detector, we will have to move the detector whenever changes since will M di change. Part 2 Part a) We can use the equation we derived in part 1a) to get: f1 d01 35mm × 100mm di1 = = ≈ 53.8mm (8) d01 − f1 100mm − 35mm Using our answer to part 1b) we have: f1 35mm M1 = = ≈ −0.54 (9) f1 − do1 35mm − 100mm Part b) First, lets derive an expression for the position image formed by the second lens. From the thin lens di2 equation we have: f2 d02 di2 = (10) d02 − f2 We want to express in terms of known variables. From the diagram, we can see that the distance of the do2 image formed by the first lens to the second lens is given by. Plugging into the above do2 = L − di1 expression we have that: f2 (L − di1 ) di2 = (11) (L − di1 ) − f2 Using our values from part a) we have that: 35mm×100mm 150mm × (300mm − ) 100mm−35mm di2 = ≈ 384mm (12) 35mm×100mm (300mm − ) − 150mm 100mm−35mm For the magnification we get:  di2 f2 (L − di1 ) 1 M2 = − = − × (13) do2 (L − di1 ) − f2 L − di1 f2 150mm ∴ M2 = − = − ≈ −1.56 35mm×100mm (L − di1 ) − f2 (300mm − ) − 150mm 100mm−35mm Part c) We can think of the overall magnification as the product of the magnification of the first lens M1 followed by the magnification of the second lens. Algebraically this means: M2 di1 di2 di1 f2 Mtot = M1 × M2 = − × − = × (14) do1 do2 do1 (L − di1 ) − f2 This specific expression will be useful for part e). For the numerical answer, we can use the product of the two magnifications found before: 35mm 150mm Mtot = M1 × M2 = × − ≈ 0.84 (15) 35mm×100mm 35mm − 100mm (300mm − ) − 150mm 100mm−35mm Since the magnification is positive, the image will be upright. Part d) From part 1a), we know that di1 → ∞ when do1 = f1. We can then evaluate the expression for as di2 di1 → ∞. f2 (L − di1 ) di2 = (16) (L − di1 ) − f2 ∴ lim di2 = f2 = 150mm di1 →∞ Part e) Looking at our previous expression for Mtot in terms of , , L di1 do2 and the focal lengths of the lenses we have: di1 f2 Mtot = × (17) do1 (L − di1 ) − f2 Using do1 = f1 and taking di1 → ∞ we get: di1 f2 f2 150mm lim Mtot = × = − = − ≈ −4.29 (18) di1 →∞ −di1 f1 f1 35mm In this case, since the magnification is negative, the image will be inverted. As we can see from the above expression, the final magnification does not depend on. This is very useful whenever you want to relay an L image over long distances or a distance longer than the sum of the focal length of both lenses. This also means that, once we set the detector at a distance from the second lens, we can change as much as f2 L we want and we won't have to move the detector. Placing an object a focal length away from a lens will yield "collimated" light at the other side of the lens, that is, rays that are perfectly parallel to the optical axis (though this is not actually true in practice due to conservation of optical extent and Gaussian wave propagation).