Chapter 18: Ray Optics Questions & Problems

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1 Chapter 18: Ray Optics Questios & Problems c h s θr= θi 1siθ 1 = 2si θ 2 = θ c = si ( ) + = m = = v s s f h s 1 Example 18.1 At high oo, the su is almost directly above (about 2.0 o from the vertical) ad a tall redwood tree casts a shadow that 10.0 ft log. How tall is the redwood tree? Solutio This problem employs the fact that rays travel i straight lies. At high oo, the su shies from its ceter, 2 o from the vertical axis, so that a triagle ca be made from the picture o the right. Apply basic trig, the taget of 2 o gives 10 ft 10 ft ta 2.0 = h = = 290 ft = h h ta 2.0 where we have rouded to two sigificat figures. Whe the su is less tha 45 o from the vertical, we expect the shadow to be shorter tha the height of the object ad loger tha the height if the su is more tha 45 o. Example 18.2 You shie your laser poiter through the flat glass side of a rectagular aquarium at a agle of icidece of 45 o. The idex of refractio of this type of glass is a) At what agle from the ormal does the beam from the laser poiter eter the water iside the aquarium? b. Does your aswer to part a deped o the idex of refractio of the glass? P Prepare: We will apply Sell s law at each iterface ad the come to a iterestig coclusio (that we might expect). si θ = si θ Solve: Use subscripts a, g, ad w for air, glass, ad water, respectively. Remember, all agles are measured from the ormal.

2 (a) si θ si 45 θ = = = g w a a si si 27 g si θ si 27 θ = = = g g si si 32 w So the laser eters the water at a agle of 32 from the ormal. (b) No, the result does t deped o the idex of refractio of the pae of glass. Try the calculatio directly from air to water to compare. 1 asiθ a si 45 θw = si = si 32 = w asi θa= gsi θg gsi θg= wsi θw, Aother way to see this is that sice ad the it must be true (if two thigs are each equal to a third thig the they must be equal to each other) siθ = si θ. that a a w w Assess: A flat pae of glass does othig to chage the agle of the light eterig the water, but it does slightly displace the ray laterally. Just as the idex of refractio of the glass does t matter, either does the thickess of the glass, but the amout the ray is displaced is affected by the thickess. Example 18.3 A ray of light travelig through air ecouters a 1.2-cm-thick sheet of glass at a 35 agle of icidece. How far does the light ray travel i the glass before emergig o the far side? ( P Prepare: Kowig Sell s law 1si θ1= 2si θ2) ad a little trigoometry we ca solve this problem. The followig figure shows the details ad helps us decide how to approach the problem. Solve: Kowig the idex of refractio of air ad glass ad the agle of icidece i air, we ca use Sell s law to determie the agle of refractio i glass. From Sell s law: AsiθiA= GsiθrG or θrg = si [( A/ G ) si θia] = si [(1.0/1.5) si 35 ] = 22.5 Kowig the agle of refractio i glass ad the thickess of the glass, it is a matter of straightforward trigoometry to obtai the distace the light travels i glass. cos θ rg = tr / r = t/cosθ or r G = 1.2 cm/cos 22.5 = 1.3 cm. Assess: Lookig at the figure, we kow that the distace should be a little more tha the thickess of the glass ad it is.

3 Example 18.4 The glass core of a optical fiber has idex of refractio The idex of refractio of the claddig is What is the maximum agle betwee a light ray ad the wall of the core if the ray is to remai iside the core? P Prepare: Use the ray model of light. We wat the ray of light to remai i the core ad ot refract ito the claddig. For a agle of icidece greater tha the critical agle, the ray of light udergoes total iteral reflectio. The critical agle of icidece is give by Equatio Solve: 1 claddig θc = si = si = 67.7 core 1.60 Thus, the maximum agle a light ray ca make with the wall of the core to remai iside the fiber is = Assess: We ca have total iteral reflectio because core > claddig. Example 18.5 A object is 15 cm i frot of a covergig les with a focal legth of 10 cm. Use ray tracig to determie the locatio of the image. What are the image characteristics: (i) real or virtual?, (ii) upright or iverted?, (iii) smaller, larger or same?, ad (iv) the image locatio? Solutio Use ray tracig to locate the image. The followig figure shows the ray-tracig diagram usig the steps of Tactics Box You ca see from the diagram that the image is i the plae where the three special rays coverge. The image is located at s = 30 cm to the right of the

4 covergig les, magificatio is more tha 1, ad is iverted ad real. Ray tracig must be doe to scale to obtai useful aswers. Example 18.6 A light bulb is 60 cm from a cocave mirror with a focal legth of 40 cm. A 5-cm-log mascara brush is held upright 20 cm from the mirror. Use ray tracig to determie the locatio of the image. What are the image characteristics: (i) real or virtual?, (ii) upright or iverted?, (iii) smaller, larger or same?, ad (iv) the image locatio? P Prepare: Two rays are eeded to determie the locatio of the image. A ray icidet parallel to the optic axis will be reflected back through the focal poit ad a ray icidet through the focal poit will be reflected back parallel to the optic axis. Real images are formed whe the reflected light actually passes through the site of the image. Solve: The figure shows that the image is at a distace of 30 cm from the mirror. Sice it has a orietatio opposite that of the object, it is iverted. The image is real sice the reflected rays actually pass through the image (we could see the image o a scree placed at this locatio). Fially, sice the image is smaller tha the object, the magificatio is less tha oe. Assess: While ray tracig is relatively simple, it ca yield a lot of iformatio. Example 18.7 A 2.0-cm tall object is 15 cm i frot of a covergig les that has a 20 cm focal legth. Where is the image located ad what is the height of the image? What are the characteristics of the image (real or virtual, upright or iverted, elarged or reduced)? P Prepare: Assume that the covergig les is a thi les ad Equatio is applicable. Solve: Usig the thi-les formula, = + = s s f 15 cm s 20 cm s = 60 cm s = 60 cm The image height is obtaied from s 60 cm M = = =+ 4 s 15 cm h = Mh = 4h = 4(2.0 cm) = 8.0 cm. Thus, the image is four times larger tha the object or The image is upright.

5 We also show later the ray-tracig diagram. The three special rays after refractig do ot coverge. Istead the rays appear to come from a poit that is 60 cm o the same side of the les as the object, so s = 60 cm. The image is upright ad has a height of 8.0 cm. Assess: The ray-tracig diagram cofirms that the positio ad height obtaied above are correct. Example 18.8 A 1.0-cm tall object is 60 cm i frot of a divergig les that has a -30 cm focal legth. Where is the image located ad what is the height of the image? What are the characteristics of the image (real or virtual, upright or iverted, elarged or reduced)? P Prepare: Assume that the divergig les is a thi les ad Equatio applies. Solve: Usig the thi-les formula, = = = s f s 30 cm 60 cm 20 cm s = 20 cm Assess: The ray-tracig diagram cofirms that both the positio ad height obtaied previously are correct. The image height is obtaied from s 20 cm 1 M = = = = 0.33 s 60 cm 3 h = Mh = (0.33)(1.0 cm) = 0.33 cm, Thus, ad the image is upright because M is positive. We also show the ray-tracig diagram i the followig figure. After refractio from the divergig les, the three special rays do ot coverge. However, the rays appear to s = 20 cm. meet at a poit that is 20 cm o the same side as the object. So The image is upright ad has a height of 0.33 cm.

6 Example 18.9 A 3.0-cm tall object is 15 cm i frot of a covex mirror that has a -25 cm focal legth. Where is the image located ad what is the height of the image? What are the characteristics of the image (real or virtual, upright or iverted, elarged or reduced)? P Prepare: We ll first use the thi-les equatio to fid the image positio ad hh / = sh /. the we ll use s = 15 cm, f = 25 cm, h = 3. 0 cm. We are give ad Solve: = = = cm s f s 25 cm 15 cm s = 1/ cm = 9. 4 cm. So So the image is 9.4 cm behid the mirror. s 9. 4 cm h = h = 3. 0 cm = 1. 9 cm s 15 cm Assess: A quick ray-tracig diagram cofirms that both the positio ad height give are probably correct.

7 Example A 3.0-cm tall object is 45 cm i frot of a cocave mirror that has a 25 cm focal legth. Where is the image located ad what is the height of the image? What are the characteristics of the image (real or virtual, upright or iverted, elarged or reduced)? P Prepare: The object distace, image distace, ad focal legth are related 1/ 1/ 1/. by s+ s = f The magificatio is related to the object ad image distace ad height m= hh / = ss /. by Solve: The image positio is obtaied by = = = s f s 25 cm 45 cm 1125 cm s = 56 cm. or The image height is obtaied by h = hss ( / ) = (3.0 cm)(56 cm/45 cm) = 3.7 cm Assess: First ote that the object is betwee the focal legth ad the radius of curvature of the cocave mirror. This tells us that the image will be iverted (opposite orietatio as the object), hece a egative image height; real (positive sig o the image distace); formed beyod the radius of curvature of the mirror; ad have a magificatio with a magitude greater tha oe (the image appears to be taller tha the object). The give iformatio regardig images formed by a cocave mirror is cosistet with the calculated values.