Question

Derive an expression for lateral shift and normal shift. on what factors these depend.

Answer 1

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.

Answer 2

Answer:

Lateral shift can be defined as the perpendicular distance of the incident ray and the emergent rays. Normal shift may be described as the apparent shifting of the position of object  during refraction of light. In both the shifts, the refractive index is the main factor of dependence.  Normal shift expression can be derived by the following formula:-

S = h(1-1/r) [S=Normal shift]

Lateral shift expression can be derived by the following formula:-

D={t/cos r} x sin (i-r)