Saved Bookmarks
| 1. |
Define normal shift. Obtain an expression for normal shift when an object in a denser medium is viewed from a rarer medium. |
Answer» Solution :The apparent shift in the position of the object kept in one medium and viewed normally through another medium.  OP = INCIDENT ray PQ = refracted ray produced backwards meet ON at I. NN. = normal OI = normal shift `/_RPO = i = ` angle of incidence = NPO `/_ QPS =r=` angle of REFRACTION= NIP t = thickness of the optical medium. ` sin i = ( NP )/( OP )`[ From the `Delta NPO`] `sin r = ( NP )/( IP )`[ From the `Delta NP]` `n_(1) sin i = n _(2) sin r ` Here, `n_(1)= n , n_(2) = 1 ` `n= ( NP )/( OP) = 1( NP )/( IP )` `n = ( NP )/( IP ) xx ( OP )/( NP) = ( OP )/( IP ) = ( ON )/(IN ) ` `:.`P is the point which is very close to N `n = ( "real DEPTH ")/( "apparent depth")` Normal shift `= OI = ON -IN` `= ON - ( ON )/( n )` `NS = ON [ 1- (1)/( n )] =t [ 1- ( 1)/ (n ) ]` Normal shift `= t [ 1- ( 1)/( n ) ]` |
|