Question

. (a) What is called lateral shift? Derive an expression for it​

Answer 1

Explanation:

When a ray of light is incident obliquely on a parallel. sided glass slab the emergent ray shifts laterally . The perpendicular distance between the direction. of the incident ray and emergent ray is called. “lateral shift''.

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Answer 2

Answer:

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+tμx+t

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

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

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

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

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

see 2nd figure, here d is lateral shift.

from ∆ABC,

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

$d=ABsin(i-r) =AB(sini.cosr-cosi.sinr) \\ = \frac{t}{cosr}(sini.cosr-cosi.sinr) \\ = t(sinr-cosi.tanr)t(sinr−cosi.tanr) .....(1)$

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),

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

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.

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