"MEASURING INDEX REFRACTION in MEDIUM from FOCAL LENGTHS. A diagram of the Problem is shown below. Surface is represented by 's'. · s · R1 radius at center C1 ray · °s· s ° · R2 radius at center C2 · ° s ° · s ° · V1 vertex of sphere 1 · ° s R1 °s · ° · V2 vertex of sphere 2 o······o·····sV1·······V2s····°··o········o S source point S C2 ° ·s n[l]s C1 P P image point s° ·R2 s S[O] Source distance n[m] s ° ·s S[I] Image distance s |V2-V1|<< S[O] or S[I]. |--------S[O]------|-----------S[I]-------| (c) Copyright PCSCC, Inc., 1993 Sign Convention: S[O] + means its left of vertex V2 focal_length < 0 S[I] + means its right of vertex V1 means R? + means center C is right of V diverging lens. *** Answer(s) to problem *** (a) First set F=60, N[L]=1.6, N[M]=1, R1=-1 (arbitrary) and R2=0. Next type (space) R2=radius_2 (enter) to set R2 to its proper radius. Finally, set F=240 and N[M]=0. Result is INDEX_M equals 1.39. ||A bi-convex lens of index 1.6 has a focal length of 60 cm in air and 240 cm when immersed in a transparent fluid. (a) What is the index of refraction of the transparent fluid? Type comma key to see answer. Type (F2) to return to application file."
