Question

normal shift derivation​

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: ANSWER \: \: } \mid}}}}}$

$\: \: \: \: \: \: \: \text{In a situation, when an object placed } \\ \text{in one medium is viewed from another} \\ \text{ medium along normal to the surface,} \\ \text{there is an apperent shift in the position } \\ \text{of the object. It is called normal shift. }$

As seen in the diagram,

$\: \: \: \: \: \tt{An \: \: object \: \: is \: \: placed \: \: at \: \: point \: \: suppose, \: 'O' , } \\ \tt{plane \: \: surface \: \: 'CD' \: forms \: \: its image \: (which \: \: is \:} \\ \tt{virtual) \: \: at \: \: 'I1' . \: This \: \: image \: \: acs \: \: as \: \: objects \: } \\ \tt{for \: \: the \: \: refraction \: \: at \: \: the \: \: surface \: \: 'EF' . } \\ \tt{Which \: \: finally \: \: forms \: \: an \: \: image \: \: at \: \: 'I' \: (virtual). } \\ \tt{The \: \: distance \: \: 'OI' \: \: is \: \: called \: \: the \: \: Normal \: \: Shift.}$

$\: \: \: \: \: \: \: \: \: \: \text{ Normal shift is directly proportional to } \\ \text{thickness of slab and refractive index of glass } \\ \text{slab.}$