TEAMS National Competition Middle School Version Photometry 25 Questions Page 1 of 13
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 = 1 d 0 + 1 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 13
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 1.501 Diamond 2.419 Carbon Disulfide 1.628 Fluorite 1.434 Carbon Tetrachloride 1.461 Fused Quartz 1.458 Ethyl Alcohol 1.361 Gallium Phosphide 3.50 Glycerin 1.473 Crown Glass 1.52 Water 1.333 Flint Glass 1.66 Air ~1.000 Ice 1.309 Carbon Dioxide ~1.000 Polystyrene 1.49 Sodium Chloride 1.544 Originally from Physics for Scientists and Engineers by Serway & Jewett (2010) 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: Page 3 of 13
Light Ray n 1 sin(θ 1 ) = n 2 sin (θ 2 ) 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 central maximum of the other (Figure 3). Page 4 of 13
Figure 3: Viewing Light as a Graph and Rayleigh s Criterion Retrieved from: http://www.kshitij-iitjee.com/study/physics/part7/chapter38/30.jpg 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 380 450 nm 450 495 nm 495 570 nm 570 590 nm 590 620 nm 620 750 nm Ref: Thomas J. Bruno, Paris D. N. Svoronos. CRC Handbook of Fundamental Spectroscopic Correlation Charts. CRC Press, 2005. Page 5 of 13
Useful Equations The Lens Equation 1 f = 1 d 0 + 1 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 = 3.281 ft = 39.37 inches (in) 1 foot = 12 inches (in) = 30.48 cm = 1/3 yard (yd) 1 mile = 5,280 ft = 1.609 km 1 lb = 4.448 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 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 Page 6 of 13
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 Commented [KR1]: Changed to 18 if they just use: M=-di/do -2/3 = -12/x x = 18 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 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 Commented [KR2]: I think your answer to 6 is incorrect: M = di/do = hi/ho di/do = 50/5 = 10 = M M = hi/ho hi = M*ho = 10*ho Height of image = 10* height of object? Page 7 of 13
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) -2.19 c) -2.88 d) 6.70 e) -6.70 9. If we have a lens which creates a reflected image 30 cm away on one side of an object 10 cm away on the other side, which material produces the Coddington shape factor with the largest magnitude? a) Ice b) Flint Glass c) Crown Glass d) Cubic Zirconia e) Quartz 10. 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 11. 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 13
12. A beam of light travels through a slab of crown glass and enters a second slab of crown glass with an angle of incidence of 26 degrees. What is the angle of refraction? a) No refraction, 0 degrees b) 26 degrees c) 64 degrees d) -26 degrees e) 154 degrees 13. 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) 17.54 degrees b) 31.47 degrees c) 29.91 degrees d) 56.05 degrees e) 330.09 degrees 14. 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 2 15. 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 21.16 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 13
For problems 16 through 17, 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) 10 8.739494118 20 17.41362033 30 25.94447977 40 34.22465021 50 42.08936521 60 49.26819412 70 55.3087574 80 59.50876401 16. 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 17. 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) 0.875 b) 1.143 c) 1.006 d) 0.921 e) 1.000 For problems 18 through 20, 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. Page 10 of 13
30 n 1 = 1 0.5 cm 1 cm α 1 α2 n 2 = 1.5 n 3 = 2 Reference 5 cm 18. 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 19. 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 20. What is the angle of refraction as the laser beam enters the third material? a) 13.23 degrees b) 19.47 degrees c) 53.30 degrees d) 14.48 degrees e) 30.00 degrees Page 11 of 13
21. 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) 1.801 x 10 5 b) 1.801 x 10 7 c) 3.602 x 10 5 d) 3.602 x 10 7 e) 9.15x 10 9 22. Which color of light is associated with the smallest angular resolution? a) Orange b) Green c) Red d) Blue e) Yellow 23. Suppose a telescope has an angular resolution of 1.342 x 10-5 radians. If the aperture diameter is 5 cm, then what is the color light we are able to view best? a) Orange b) Green c) Red d) Blue e) Yellow For problems 24 through 25, refer to the following plot of two diffraction patterns for separate images in the same view of a telescope. Page 12 of 13
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 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 13 of 13