Wellifthevelocityoflightis c whenittravelsthroughairthenforthevelocityonceittravelsintheglassisequalto c dividedbytheindexofrefraction.
Sointhiscaseweknowthat c is 3 times 10 tothe 8 meterspersecondandthenwedividethatby 1.6 andthatwillgiveusthenthenewvelocityoflight.
And I justremembered I didnotbringmycalculatorsoletmegogetmycalculatorrealquickand I'llberightback.
Nowthat I havemycalculatorlet's findouthowfastlighttravelsthroughglass, wellforglassthathasanindexofrefractionof 1.6 sowetake 3 dividedby 1.6 andweget 1.875 sothisisequaltoandletmegolikethisequalto 1.875 times 10 tothe 8 meterspersecondsoit's a significantslowdownstillfastbutyoucanseethatlightdoestravelat a differentspeedandthere's a mediumotherthanairorjustfreespace.
Sowhathappensnowwhen a rayoflighttravelsfromonemediumtoanotherandherewehaveairandwater, doesn't matterjusttwodifferentmediumsandtheindexofrefractionforairand 1 equals 1 forwaterand 2 equals 1.33 Nownoticethatthelightdoesnottravelstraightacrosstheboundaryat a linethat's perpendicularornormaltothesurface, ittravelsatananglerelativetothenormalandwhenitcrosses a boundarylikethatsomethingelsehappenstothelightbesidesitslowingdown, italsochangesdirection, itbendsoritrefractsaswecallitandinthecaseoflightthattravelsfromanindexofrefractionwhichissmallertoanindexofrefractionwhichislargerthebendingwillbewhatwecalltowardsthenormalsotheanglebetweenthenormalandtheraywillbecomesmallerandsothelightwilldosomethinglikethisandthiswillnowbethetasub 2 noticethatthetasub 2 isindeedsmallerthanthetasub 1 Wellhowdoweactuallyfindoutwhatthoseanglesare?
Let's do a quickexamplelet's sayforexamplethattheta 1 isequalto 30 degreesthenwhatdoyouthinktheta 2 willbeequalto?
WellitturnsoutthatSnell's lawenablesustodothatandSnell's lawsaysthat n1 timesthesineoftheta 1 equals n2 timesthesineofthetasub 2 andagainwhatwe'retryingtofindiswe'retryingtofindtheanglesineofthetasub 2 whichmeanswehavetoalgebraicallysolveforthatanglesowe'refirstgoingtofliptheequationaroundsowecansaythat n2 timesthesineoftheta 2 equals n1 timesthesineofthetasub 1 thenthenextstepiswe'regoingtodividebothsidesby n sub 2 sowehavethesineofthetasub 2 isequalto n1 over n2 timesthesineofthetasub 1 andfinallytofindthetasub 2 wetakethearcsinesothetasub 2 equalstheinverseorarcsineofthiswholethingrightherewhichis n1 over n2 timesthesineofthetasub 1 nowpluginthevaluesthatwehavesoweknowthatthetasub 1 is 30, n1 is 1 and n2 is 1.33 inthisparticularcasebecauseit's fromairintowaterwecansaythatthetasub 2 isequaltothearcsineof n1 whichis 1, n2 whichis 1.33 andtimesthesineof 30 degreesokaylet's findoutwhatthat's equaltosoyoutakethesineof 30 thatshouldbe 0.5 dividethatby 1.33 andthentakethearcsineofthatandweget 22.1 degreessothetasub 2 isequalto 22.1 degreesandthat's whatwecalltheangleofrefractionlet's writethatdownsothat's calledtheangleofrefractionokaysonoticethatwhentheangleis 90 degreesor I shouldsay 0 degreesbetweenthenormalandtheraythere's nobendingoflightthelightgoesstraightthroughtheonlythingthathappensisthatitslowsdownrelativetotheindexofrefractioninthiscaseletmejustputdown v sub 2 andthat's thevelocityinthesecondregionisdependentupontheindexofrefractioninthesecondregionbutiftheraycomesinatananglethennotonlydoesitslowdownonceitcrossestheboundaryitalsochangesdirectionithas a newdirectionthedirectionisdifferentfromtheangleofincidenceandusingSnell's lawishowwefindwhatthatangleisalrightsothere's a nicelittleintroductiontotheconceptofrefraction