TEAMS National Competition High School Version Photometry 25 Questions

Transcription

1 TEAMS National Competition High School Version Photometry 25 Questions Page 1 of 14

2 Telescopes and their Lenses Although telescopes provide us with the extraordinary power to see objects miles away, the mathematics behind the technology is based on a set of simple equations. We can begin by discussing a refracting telescope, which contains two main lenses, a larger one called the objective lens and a smaller eyepiece meant for the user to look though. The telescope functions using the properties of the lenses to manipulate light and deliver it to our eyes in a magnified form. One property is the underlying shape; for telescopes, a lens can be convex or concave. A convex lens is thick in the middle and thin as we approach the edges while a concave lens is thinner in the middle and thicker around the edges. In order for the refracting telescope to perform as we would like, the objective lens must be convex. Based on the geometry on the surface, the light rays will converge at a point some distance f away, called the focal distance. On the other hand, a concave lens will cause the light rays to diverge. (Fig 1.) Figure 1: Convergent and Divergent Lenses For convergent lenses, an object some distance d 0 away from the lens will have a reflected image a distance d i on the other side. The three quantities of interest are related through the lens equation. 1 f = d 0 d i The left hand side of the lens will be considered the front (the virtual side) and the right hand side will be the back (the real side). This would mean the sign of d 0 would be positive if the object is to the left of the lens and d i would be positive if the image is to the right of the lens. Likewise, if the object is on the right hand side of the lens, then the sign of d 0 would be negative and if the image is on the left, d i would be negative. The focus is also signed; the right-hand side is considered positive for the focus while the left-hand side is negative. Table 1 provides a summary of the sign convention. Page 2 of 14

3 Table 1: Summary of Lens Sign Convention Positive When Negative When Object s distance d 0 In front of mirror / to the left In the back / to the right Image s distance d i In the back / to the right In front of mirror / to the left Focus f Lens is convergent / convex Lens is divergent / concave Image Height Image is upright Image is inverted The geometry of the lens will also aid in defining the magnification. The reflected image of the object with height h 0 will have a scaled height h i. We can define the magnification M using the distances away from the lens as well: M = h i h 0 = d i d 0 Based on the signs, we can tell if the image is upright (positive) or inverted (negative). Not all lenses are perfect, increased refraction (or bending) of light can cause an imperfection in the reproduced image this is effect is called spherical aberration. Figure 1 displays an ideal lens; that is, a lens lacking spherical aberration. Practically, all lens will suffer from this effect, but we are able to minimize it using the Coddington shape factor, C. C = 2(n2 1) n + 2 (d i + d 0 d i d 0 ) where n is the index of refraction a number describing how light propagates through a medium. Table 2 contains the index of refraction for a select few common materials. Table 2: Index of Refraction for Common Materials Material Index of Refraction (n) Material Index of Refraction (n) Cubic Zirconia 2.20 Benzene Diamond Carbon Disulfide Fluorite Carbon Tetrachloride Fused Quartz Ethyl Alcohol Gallium Phosphide 3.50 Glycerin Crown Glass 1.52 Water Flint Glass 1.66 Air ~1.000 Ice Carbon Dioxide ~1.000 Polystyrene 1.49 Sodium Chloride Originally from Physics for Scientists and Engineers by Serway & Jewett (2010) Page 3 of 14

4 Refraction between two materials, beyond ordinary lenses, can be computed using Snell s Law, which relates the indices of refraction for each material and how they impact the material. Snell s Law is stated as follows: n 1 sin(θ 1 ) = n 2 sin (θ 2 ) Light Ray n 1 θ 1 θ 2 n 2 Normal Refracted Ray Figure 2: Visual Representation of Snell s Law where θ 1 is called the angle of incidence and θ 2 is called the angle of refraction. Each angle is measured clockwise from the normal, a line which is perpendicular to the surface of the material. A refracting telescope uses this fundamental idea to create images by using two lenses, one called the objective lens and the other is the eyepiece lens. To achieve the best detail when viewing images, we must consider Rayleigh s criterion. In order to distinguish between two patterns, in the worst case, the first minimum of one image must fall at the same point as the central maximum of the other (Figure 3). Page 4 of 14

5 Figure 3: Viewing Light as a Graph and Rayleigh s Criterion Retrieved from: From Figure 3, we can see that if the patterns overlap too much, then the images cannot be distinguished from one another. This criterion is often formulated as θ = 1.22 λ D where θ is the angle of resolution, D is the diameter of the aperture, and λ is the wavelength. For reference, a table of wavelengths for the visible color spectrum are included in Table 3. Table 3: Wavelengths for the Visible Color Spectrum Color violet blue green yellow orange red Wavelength nm nm nm nm nm nm Ref: Thomas J. Bruno, Paris D. N. Svoronos. CRC Handbook of Fundamental Spectroscopic Correlation Charts. CRC Press, Page 5 of 14

6 Useful Equations The Lens Equation 1 f = d 0 d i Magnification M = h i h 0 = d i d 0 Coddington Shape Factor C = 2(n2 1) n + 2 (d i + d 0 d i d 0 ) Snell s Law n 1 sin(θ 1 ) = n 2 sin(θ 2 ) Rayleigh s Criterion θ = 1.22 λ D Conversions 1 meter = 100 cm = 1,000 mm = ft = inches (in) 1 foot = 12 inches (in) = cm = 1/3 yard (yd) 1 mile = 5,280 ft = km 1 lb = N Photometry Questions 1. When an upright object is placed between the focal point of left side of a converging lens, how can the resulting image be described? a) Real, upright, to the left b) Virtual, upright, to the left c) Real, inverted, to the right d) Virtual, inverted, to the right e) Real, inverted, to the left Page 6 of 14

7 2. An image is formed by a convex lens 50 cm to the right of the lens. The object is placed 10 cm to the left of the lens. In this configuration, the focal distance is most nearly a) 8 cm to the left b) 8 cm to the right c) 12.5 cm to the right d) 12.5 cm to the left e) Focus too far away to be seen 3. Consider a concave lens where the reflection of an object is 7.5 cm to the left of the lens with a magnification of 3/4. Find the distance of the object from the lens. a) 10 cm to the left b) 10 cm to the right c) 15 cm to the left d) 15 cm to the right e) 7.5 cm to the left 4. Say you have a concave lens with a focal distance of 12 cm and a magnification of -2/3, find the distance of the reflected image. a) 30 cm to the left b) 30 cm to the right c) 20 cm to the left d) 20 cm to the right e) 18 cm the right 5. A convex lens has a focal length with magnitude 30 cm. Where is the image of an object that is placed 30 cm left of the lens? a) 15 cm to the right b) 15 cm to the left c) 20 cm to the left d) 0 cm, reflected back onto object e) Image too far away to be seen 6. An object is positioned 5 cm to the left of a converging glass lens which reflects an image 50 cm to the right of the lens. What does this tell you about the image itself in terms of size and orientation? a) Shrunk and upright b) Shrunk and inverted c) Enlarged and upright d) Enlarged and inverted e) Unchanged Page 7 of 14

8 7. An object is positioned 5 cm to the right of a converging glass lens which reflects an image 5 cm from the left of the lens. What does this tell you about the size of the image? a) 5 times larger b) Double the size c) Unchanged d) Half the size e) Five times smaller 8. Suppose we made a lens out of crown glass. What is the Coddington shape factor if we have an object 5 cm away being reflected as an image 4 inches away from the lens on the other side? a) 2.19 b) c) d) 6.70 e) A light beam hits a cube of ice entirely normal to the surface. What is the angle of refraction? a) 0 degrees b) 90 degrees c) 180 degrees d) 270 degrees e) No refraction 10. What material M best imitates the situation pictured in the following figure.? a) Crown Glass b) Glycerin c) Cubic Zirconia d) Polystyrene e) Ice Air M Page 8 of 14

9 11. A beam of light traveling down and to the right through a block of Quartz enters a block of diamond at an angle of 30 degrees clockwise measured from the diamond to the beam. Find the angle of refraction the refracted beam makes with the normal. a) degrees b) degrees c) degrees d) degrees e) degrees 12. Given the following configuration where θ 1 < θ 2, a beam of light traveling between two mediums, which mathematical relationship must be true? θ 1 n 1 a) n 1 < n 2 b) n 1 > n 2 c) n 1 n 2 d) θ 1 + θ 2 = 90 e) θ 1 θ 2 θ 2 n A controlled beam of light is initially incident on a slab of quartz 30 degrees from the normal, but is adjusted to produce a refracted ray within cubic zirconia at an angle degrees from the normal. Approximately what amount of offset in the angle is needed at the point of incidence in order to produce the refracted ray? a) 26 degrees b) 16 degrees c) 19 degrees d) 3 degrees e) 2 degrees Page 9 of 14

10 For problems 13 through 14, refer to the following table of values obtained from adjusting the angle of incidence for a laser beam passing through two materials each of which have an index of refraction, n 1 and n 2 respectively. Angle of Incidence (deg) Angle of Refraction (deg) What do the sequence of values for the angle of refraction tell you about the change in angular position of the laser beam? a) The angle of refraction is slowly becoming a right angle b) The beam is gradually moving away from the normal c) The beam is approaching the normal d) The refracted beam will eventually approach an angle and not bend further e) The beam will become tangential to the surface of the second material 15. We can verify Snell s Law by finding the sine of the angle of incidence, x, and the sine of the angle of refraction, y, then plot y/x. What kind of graph will be plotted if we do so? a) Exponential b) Logarithmic c) Quadratic d) Linear e) Power 16. From the experiment, we can find that n 1 n 2 (n 1 is proportional to n 2 ). This means there is a number k that satisfies the equation n 1 = kn 2. Find k. a) b) c) d) e) Page 10 of 14

11 For problems 17 through 19, refer to the following diagram. A ray of light is incident upon a material at an angle of 30 degrees from the normal. It then passes through a 1 cm slab of material with an index of refraction equal to 1.5, then exits into a material with an index of refraction equal to 2. The picture is NOT to scale. 30 n 1 = cm 1 cm α 1 α2 n 2 = 1.5 n 3 = 2 Reference 5 cm 17. At what distance from the reference line does the refracted ray within the second material hit the third material? a) 0.71 cm b) 0.85 cm c) 1.06 cm d) 0.35 cm e) 1.21 cm 18. What is the relationship between the angles α 1 and α 2? a) α 1 < α 2 b) α 1 > α 2 c) α 1 = α 2 d) α 1 + α 2 = 90 e) No relationship remains the same for any α 1 and α 2 Page 11 of 14

12 19. What is the angle of refraction as the laser beam enters the third material? a) degrees b) degrees c) degrees d) degrees e) degrees For problems 20 through 22, consider 100 slabs of different material each with a different index of refraction: n 1, n 2, all the way up to n 100. The light beam enters with an angle of incidence equal to 45 degrees clockwise from the normal. Suppose the indices increase based on the following formula, n i = i, where i represents the ith index of refraction for the ith material. 20. What is the angle of refraction from the 1 st to 2 nd material? a) degrees b) degrees c) degrees d) degrees e) There is little to no refraction 21. What is the angle of refraction from the 56 th to 57 th material? (Hint: Use the following trick, n 1 sin(θ 1 ) = n i sin(θ i ) for the ith material) a) degrees b) degrees c) degrees d) degrees e) There is little to no refraction 22. What can you say about the final angle of refraction? a) The angle of refraction is nearly the same as the angle of incidence b) The beam is further away from the normal c) The refracted beam is completely horizontal d) The beam is closer to the normal e) There is little to no refraction Page 12 of 14

13 23. What is the angular resolution for a telescope with a lens aperture radius of 1 inch to view red light for the largest wavelength? a) x 10 5 b) x 10 7 c) x 10 5 d) x 10 7 e) 9.15 x 10 9 Page 13 of 14

14 For problems 24 through 25, refer to the following plot of two diffraction patterns for separate images in the same view of a telescope. Intensity / Brightness π λ sin(θ) 24. When is the pattern the darkest for Image 1 or Image 2? a) When both curves intersect b) When both curves reach a local maximum c) When either curve reaches a local minimum d) When the difference between the curves are the largest e) When the difference between the curves are the smallest 25. Consider each rectangle width in the plot as 1 unit on the x axis. At a minimum, how much does Image 1 need to be shifted in order to resolve the resulting image? a) 3 units to the left b) 2 units to the left c) 1 unit to the left d) 1 unit to the right e) 2 units to the right Page 14 of 14