Question

Expression of normal shift

Answer 1

Normal shift: Consider the situation shown in the figure.



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

OI = (1−1μ)t

We can derive this as,

Let OA = x, then from the snell's law applied at surface CD,

 AI1 = μx 

So, the object distance for the refraction at surface EF is BI1 = μx+t. So by applying snell's law at the surface EF,

BI'=BI1μ = x + tμ

So, shift OI' is,

OI' = (AB+OA) − BI' ⇒(t+x) − (x+tμ) = t−tμ ⇒OI'= (1−μ) t

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