Question

*- Jordi 1. 14 points. You wish to project an image of a slide onto a screen for a slide show. The slide is initially in a holder 0.0500m from the lens and the screen is 3.00m from the lens. The image is well-focused on the screen. a. Determine the focal length of the lens used to project the image. b. Is the lens converging or diverging? How do you know? c. If you instead want to use a screen 2.00m away from the same lens, what should the distance between the lens and slide be? d. Which setting (2.00m screen or 3.00m screen) will produce the larger image of the slide? e. Should the slide be put in the holder right side up or upside down if you want the image on the screen to be right side up? 2. n, sino = nasin! 13 Points, Light strikes an interface between plastic ( 1.5) and glass ( 2.3) as shown in figure 2a. The angle of incidence of the ray is 25 degrees. Parts a and b address this ray. Part addresses the two incident rays in figure 2b, which travel toward an identical interface from inside the glass. a. Calculate the angle of reflection of the ray in 2a. b. Calculate the angle of refraction of the ray in 2a. C. Draw all rays leaving the interface (reflected and refracted) in figure 2b i. From incident ray I ii. From incident ray II. d. For all rays drawn in part cabove, calculate the angle of reflection or refraction 3. Coherent light of wavelength 600nm shines on two narrow slits separated by a distance 2.00mm The light then shines on a screen located 3.5m behind the slits. a. Sketch the brightness pattern observed on the screen. b. Determine the distance of the second bright spot from the center of the screen. c. Determine the distance of the third dark spot from the center of the screen. 4. A particular person's eye has a near point of 40.0cm. An optometrist prescribes contact lenses which will allow the person to view objects as close in as 25cm away. a. Determine the focal length of the contact lens. b. Briefly explain how the contact lens works to "fix" the person's near point

Answer 1

distance of the slide holder from lens (do)= 0.05m distance of seen from lens (di) =3 m. ca) using thin lens equation de I+Los f 3 0.05 I - 1 tt do = tt I -> f = 18.04927 B) The given lens is a converging clens as only converging lens com forces the days ontre a screen, thus forming real and inverted image Again wing thin lens equation, do I Ž do » do= [o-osoum (d) Magnification in first case = 3 = 60 magnification in second Cose = 2 ou 39.7 Screen at a distance of 3 m produces a larger magnified image e) Slide shoould be inserted in the holder- "upside down" because converging lens forms inverted images of the object on the screen f di 0.0492 o.os . G.Osou

- 2mm D = 3.5m d = 6oonm a) » Intensity pattern in di d=6oonm ED - Screen (b) Kuand (For bright spot) n = xx600x109x3.5 xx10-3 Comu). To (① - 2.1810-3 m or 2.1mm (2n-1)dD (dark shot) ad = (2(3) -1) 600X10-9x3.5 kn xxxx 10-3 = 2.625 x 10-3 m or (2.625 mm

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