Saved Bookmarks
| 1. |
A solid glass sphere with radius R and an index of refraction 1.5 is silvered over one hemisphere. A small object is located on the axis of the sphere at a distance 2R to the left of the vertex of the unsilvered hemisphere. Find the position of final image after all refractions and reflections have taken place. |
Answer» Solution :The ray of light FIRST gets refracted then reflected and then again rcfracted. For first REFRACTION and then reflection the ray of light travels from left to right while for the last refraction it travels from right to left. Hence, the SIGN convention will change accordingly.  First : refraction Using ` (mu_2)/(v ) (mu_1)/( u) = (mu_2-mu_1)/(R ) ` withpropersign convertions , we have`(1.5 )/(v ) -(1.0 )/( -2R) =(1.5-1.0 )/(+R)thereforev_1 = oo` Second : reflection Using ` (1)/(v ) +(1)/(u )=(1)/(f)=(2)/(R ) ` withpropersign conventions we have ` (1)/( v_2) +(1)/(oo) =- (2)/(R )` ` thereforeV_2=- R/2 ` Third: refraction Again using`(mu_2 )/(v ) -(mu_1)/(u )= (mu_2 - mu_1)/(R ) ` withreversedsignconvention, we have ` (1.0)/(v_3) - (1.5 )/( -1.5R) =(1.0 -1.5)/(-R)` ` or V_3 =- 2 R ` i.e., FINALIMAGE isformedonthe vertexof thesilveredface |
|