Question

what do you mean by normal shift obtain its expression​

Answer 1

Explanation:

See first figure. here OI' is normal shift of object.

Let OA = x,

then from Snell's law,

AI1 = \mu x

so, the object distance for the refraction at surface EF is BI1 = \mu x+t

so, applying Snell's law at the surface EF,

BI' = \frac{BI_1}{\mu}=\left(x+\frac{t}{\mu}\right)

so, shift OI' = (AB + OA) - BI'

= (t + x) - \left(x+\frac{t}{\mu}\right)

\boxed{\bf{OI'=t\left(1-\frac{1}{\mu}\right)}}

see 2nd figure, here d is lateral shift.

from ∆ABC,

AB=\frac{AC}{cosr}=\frac{t}{cosr}

d=ABsin(i-r)=AB(sini.cosr-cosi.sinr)

= \frac{t}{cosr}(sini.cosr-cosi.sinr)

= t(sinr-cosi.tanr).....(1)

from Snell's law,

1sini=\mu sinr

or, sinr=\frac{sini}{\mu}

tanr=\frac{sini}{\sqrt{\mu^2-sin^2i}}.......(2)

from equations (1) and (2),

d=\left[1-\frac{cosi}{\sqrt{\mu^2-sin^2i}}\right]tsini

for small incident angle,

\boxed{\bf{d=\left[1-\frac{1}{\mu}\right]ti}}

here it is clear that, it depends on incident angle, refractive index and thickness of refractive medium.