# Position: # (your seat #)………………………………………….. Your Name: Last, First…………………………………………… Name of Workshop Instructor: Last, First……………………….. Exam: # (#2)………………………………………………………

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Position: # (your seat #)………………………………………….. Your Name: Last, First…………………………………………… Name of Workshop Instructor: Last, First……………………….. Exam: # (#2)……………………………………………………… Date: March 22 nd, 2005………………………………………… Q1 (5 points). The below figure shows a light ray being (totally) internally reflected by the slanted surface of a prism. What is the minimum value for the index of refraction of the prism for this to occur as shown? A. 1.32B. 1.42 C. 1.52D. 2.00 12345679 101112 air Q12 (C, E, F) are for bonus points. Answer: n 1 sin  1 =n 2 sin  2 = 1 Therefore, n 1 =1/sin  1 =1/sin 45°=1.414 B is correct.

Q2 (6 points). If an object is placed 8 cm in front of a converging lens with a focal length of 2 cm, its image is at a location, relative to the lens center A. 2.67 cmB. 5.3 cm C. 8 cmD. -4 cm Answer: Q3 (4 points). The image in Question 2 is A. real and erectB. real and inverted C. virtual and erectD. virtual and inverted Answer: B is correct 1/f = (1/d 0 ) + (1/d i ) Therefore, (1/d i )= 1/f - (1/d o )=0.5 – 0.125 = 0.375 d i = 2.67 cm

Q4 (3 points). A light ray is incident on a mirror, and is reflected from the mirror. The incident light ray is drawn below. What is the reflected ray? Please explain your answer. Q5 (8 points). I had trouble with the reading of a CD. I realized that the surface had some fingerprints. Please explain in a brief verbal argument what happened. How does a CD work? Answer: The compact disc, although appearing to be stationary in the tutorial, should actually be envisioned as rotating at high speed. A beam is emitted by the laser and directed onto a single track on the disc by the prism/beamsplitter. As the disc rotates, the beam encounters a series of pits and landings that determine whether the beam is reflected back into the detector (from a landing) or scattered (a pit). Incident ray A B C D mirror Answer: A is correct. The reflected ray does not go beyond the mirror, such as in B, C, and D.

Q7 (7 points). If do = 15.6 cm and f = 6.6 cm, what is the image distance to the nearest tenth of a cm? Q6 (5 points). A light ray strikes a horizontal mirror and is reflected onto a vertical mirror. If θ = 32 degrees and d = 1.85 meters, what to the nearest degree is φ? Answer: The first mirror: the incidence angle  is equal to the reflected angle. The second mirror: The incidence angle is equal to the reflected angle . The sum of  and  is 90 . Therefore,  =58 . Answer: 1/f = (1/d 0 ) + (1/d i ) Therefore, (1/d i )= 1/f - (1/d o )=(1/6.6) – (1/15.6) = = 0.1515-0.0641=0.0874 d i = 11.4 cm

Q9 (5 points). Light strikes a flat piece of glass (n=1.50) at an incident angle of 60°. What is the angle of refraction in the glass? What is the angle with which the ray emerges from the glass? Answer: The incidence angle is equal to the reflected angle. Therefore, the reflected angle is 60°. Snell’s law is n 1 sin  1 = n 2 sin  2 sin  2 =(n 1 /n 2 )sin  1 =(1/1.5)sin 60°=0.666×0.866=0.576  2 =35.1 . Q8 (3 points). The speed of light is often listed as equaling 3.00 x 10 8 m/s. Does light always travel at that speed, or can it travel faster? Can it travel slower? Why do you believe this is so. Answer: No. Yes, it can travel slower. This happens, as the light Interacts with the atoms of the material through which it goes through.

Q11 (3 points). At 600 K, the value of the wavelength for which the intensity is peaked is 5,000 nm. If the temperature is tripled (reaching 1800 K), then the wavelength for which the intensity of radiation is peaked is now A. 40,000 nmB. 2,500 nm C. 20,000 nmD. 1,667 nm E. 5,000 nm Answer: λpT=constant 5000×600 nm K= λ p 1800 nm K λp = 1,667 nm Q10 (3 points). Red light and blue light impact on retina of your eye, with roughly equal intensity. Then, your eye-brain system perceives the color of A. WhiteB. Black C. CyanD. Yellow E. Magenta Answer: E. Magenta

Q12. A beam of light travels from water into a piece of diamond in the shape of a triangle, as shown in the diagram. Step-by-step, follow the beam until it emerges from the piece of diamond. A (5 points). How fast is the light traveling inside the piece of diamond ? Answer: The speed can be calculated from the index of refraction:

B (5 points). What is  2, the angle between the normal and the beam of light inside the diamond at the water-diamond interface? C (7 points). The beam travels up to the air-diamond interface. What is  3, the angle between the normal and the beam of light inside the diamond at the air-diamond interface? Answer: A diagram helps for this. In fact, let's look at the complete diagram of the whole path, and use this for the rest of the questions. The angle we need can be found from Snell's law: Answer: This is found using a bit of geometry. All you need to know is that the sum of the three angles inside a triangle is 180°. If is 24.9°, this means that the third angle in that triangle must be 25.1°. So:

E (5 points). What happens to the light at the diamond-air interface? D (5 points). What is the critical angle for the diamond- air interface? Answer: F (5 points). The light is reflected off the interface, obeying the law of reflection. It then strikes the diamond-water interface. What happens to it here? Answer: Because the angle of incidence (64.9°) is larger than the critical angle, the light is totally reflected internally. Again, the place to start is by determining the angle of incidence,  4. A little geometry shows that: The critical angle at this interface is : Because the angle of incidence is less than the critical angle, the beam will escape from the piece of diamond here. The angle of refraction can be found from Snell's law:

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