PHYS1004 Problem Sheet - Optics - with Solutions

Transcription

1 PHYS004 Problem Sheet - Optics - with Solutions Do not write on these question sheets - there is not enough room to do a good job of answering these questions. The answers need to be written in your life physics notebook. PART - Questions to do during the second part of today s workshop. Practice drawing a ray diagram showing an object being viewed through a converging lens. Include the three principle rays. (a) Is the image formed with a converging lens real or virtual? (b) Is an image formed with flat mirror real or virtual? (a) You should understand how to draw the three principle rays although you only need two rays to define the position of the image. The image formed through a converging lens is real (rays pass through it). (b) the image formed from a mirror is virtual - the rays only appear to be coming from the image which looks like its in the mirrow - the light can behind the mirror - its bouncing off the surface. 2. An observer O, facing a mirror, observes a light source S. Where does O perceive the mirror image of S to be located? The image of a point in a mirror is always on a line perpendicular to the mirror surface and as far behind the mirror as the point is in front of it. So it is point 4. Trace the rays back to prove it to yourself!

2 3. If were given a lens with focal length f and we place an object d o away from the lens, can we predict where the image is? + F Where is the image? d o f Yes, there are 2 ways: With a pencil, a ruler, a paper - ray diagram. (Or using the thin lens equation.) There are 3 easy-to-draw rays: + d i d i is how far the image is from the lens F 2 F 3 d o f The image is where these rays intersect (real image), or where they appear to intersect (virtual). in practice, drawing two rays is enough. 4. What will happen to the image of the tree if the top half of the lens is covered? + +? The image is still fully formed, but less bright because there are less light rays forming the image. 2

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4 PART 2 - Questions to do at home or at the drop-in tute 5. A team in a large aquarium wishes to tranquillise a shark in order to run some health checks on the creature. An optical targeting system is used to aim a tranquilliser gun at the shark. The laser emits a beam with wavelength 630 nm (under the water). The refractive index of water is.33. (a) What is the speed of the light in the targeting beam above the water? (b) What is the speed of the light in the targeting beam in the water? (c) What is the frequency of the light in the water? (d) What is the frequency of the light in the air? (e) What is the wavelength of the light in the air? (a) The speed of the light above the water is the speed of light in air, c = m/s. (b) The speed of light in the targeting beam in the water would be less as water is more dense. v in water = c(n air /n water ) = (.00/.33) = m/s (c) The frequency of the light is f = v λ f = v λ = m/s m = Hz (d) The frequency of the light in the air is the same as the frequency of the light in the water. ( Hz). (e) The wavelength of the light in the air is different to the wavelength in water λ = v in water f = m/s Hz = 830 nm. 6. Arthroscopic surgery is a minimally invasive surgical procedure in which an examination and even treatment of damage of the interior of a joint is performed using a type of endoscope that is inserted into the joint through only a small incision. Endoscopes rely on the transmission of light into and out of the joint through optical fibres which effectively trap light in flexible glass fibres. Based on your understanding of Snell s Law of refraction explain how the light becomes trapped in the glass fibre but can be easily detected at the flat ends of the fibre. The light can be trapped in a medium when the angle the light makes with a medium exceeds some critical angle - and total internal refection occurs. Consider Snell s Law: n sin θ = n 2 sin θ 2 4

5 If n 2 < n the critical angle occurs when θ 2 = 90 degrees. So if we consider glass of n =.5 and air n =.00 we get an angle of 42 degrees. The light in the glass tube approaches the walls at angles of close to 90 degrees - and thus is stuck in the glass fibre. However once the light reaches the flat horizontal end of the fibre, it can easily escape and be detected. 7. Many sea-birds and other bird of prey dive under water in order to catch fish. In order to successfully catch the fish such bird must account for the physics of the light changing paths as it moves from the water to the air. If a small fish is located at depth d under the water of a still pond, what is the apparent depth of the fish as viewed by a bird flying directly above the fish? The refractive index of water is n =.33 and air n =.00. (Hint - first draw a diagram with the rays of light coming from the fish and use the small angle approximation tan θ sin θ). Consider the rays of light coming from the fish (position O) out into the air. The light will be bent way from the normal as it is going from water to air since air has a lower index of refraction than water. Thus if we trace the rays back where they appear to be coming from the image of the fish is at a different depth d (position O ). Is it clear that d <d. s θ 2 s θ d θ 2 d θ Only two rays are needed to locate the image and solve for d. One can be the ray normal to the surface and the other incident on the surface at an angle θ. This leaves the water at a slightly larger angle θ 2. n water sinθ = n air sinθ 2 To solve for d we can use the two triangles that are formed that share the same side s - the distance between the points where the rays meet the surface. From these trianges we can see that: tan θ = s d and tan θ 2 = s d Now using the small angle approximation tan θ sin θ we see that: s n water d = n s air d 5

6 Solving for the ratio of d d d d = So the apparent depth is 0.75 of the actual depth. n air n water = Draw ray diagrams for a converging lens with focal length f, where the object distance d o is (a) larger than f but smaller than 2f (b) larger than 2f Show that the magnification is greater than one in case (a) and less than in case (b). (a) magnification is greater than. (b) magnification is less than. (Both figures from Converging-Lenses-Ray-Diagrams) 9. Tim is trying to take a photo of Linda with an old film camera. Usually when he takes a photo with this camera he accidentally cuts the top off Linda s head off. This time he is determined to get a photo showing both of them smiling at the camera. Linda is standing 3.0 m away, and the camera lens has a focal length of 40 mm and the film height is 24 mm. (You should start solving this problem by drawing ray diagrams!) (a) What is the lens to film distance of this camera? 6

7 (b) What is the linear magnification of the image? (c) If Linda is just going to be completely in the photo, so that her feet are just seen at the bottom, and her head completely seen at the top, how tall is she? (a) The question is asking what is the image distance - since images are formed on the film. Using the thin lens formula + = d o d i f so = d i f = d o 0.04 = = therefore d i = = 40.5mm This makes sense - the image distance and focal length must be approximately the same, so that the camera can take photos of things far away. (b) The linear magnification of the image is M = d i = 0.04 = d o 3.0 (c) If Linda is just going to be completely in the photo, so that her feet are just seen at the bottom, and her head completely seen at the top, her height is (using the relation from lectures that M = h i /h o h o = h i M = 0.024m =.73.7m 0.04 We were only given two significant figures in the question. If we had carried three significant figures through from the previous part, we d get h o = 0.024/0.035 =.77 =.8m, which is more accurate. 7