Question

Draw a neat diagram to show lateral shift of a ray refracted through
a parallel-sided slab. Indicate the lateral shift in the diagram by a
double-headed arrow.
​

Answer 1

LATERAL SHIFT IN REFRACTION THROUGH SLAB:

$\setlength{\unitlength}{1cm}\begin{picture}(12,12)\put(0,0){\framebox(9,5)}\multiput(4,6.5)(2,-5){2}{\line(0,-1){3}}\thicklines\qbezier(4,5)(4,5)(6,0)\qbezier(1.5,7.5)(3,6)(4,5)\qbezier(6,0)(6,0)(8.5,-2.5)\multiput(4,5)(0.5,-0.5){14}{\line(1,-1){0.4}}\linethickness{0.24mm}\qbezier(3.5,5.5)(3.7,5.9)(4,5.7)\qbezier(6,-0.6)(6.3,-0.85)(6.5,-0.5)\qbezier(4,4)(4.3,3.8)(4.38,4.1)\qbezier(6,1)(5.8,1.2)(5.62,0.9)\put(2.5,6.5){\line(-1,0){0.4}}\put(2.5,6.5){\line(0,1){0.4}}\put(7.5,-1.5){\line(-1,0){0.4}}\put(7.5,-1.5){\line(0,1){0.4}}\qbezier(5,2.5)(5,2.5)(4.6,2.7)\qbezier(5,2.5)(5,2.5)(5.2,2.9)\thinlines\qbezier(8,-2)(8,-2)(9.5,-0.5)\put(3.8,6.7){\bf M}\put(3.8,3){\bf M'}\put(5.8,1.8){\bf N}\put(5.8,-2){\bf N'}\put(1,7.7){\bf A}\put(4.3,5.3){\bf B}\put(5.4,-0.5){\bf C}\put(7.5,-2.4){\bf D}\put(9.8,-0.5){\rm E}\put(0.5,5.5){\sf AIR}\put(0.5,2.5){\sf GLASS}\put(0.5,-0.8){\sf AIR}\put(3.5,6){\bf i}\put(6.4,-1.2){\bf e}\linethickness{0.24mm}\qbezier(4,3.85)(4.3,3.7)(4.4,3.9)\qbezier(6,1.15)(5.8,1.3)(5.6,1.1)\put(5,5.5){\sf DE=Lateral Displacement}\end{picture}$

• Here, you can see that a light ray incident on a rectangular slab is undergoing refraction twice at the two parallel surfaces.

• At the 1st interface, the ray refracts towards the normal.

• At the 2nd interface, the ray refracts away from the normal.

• This results in the final emergent ray to be parallel to the incident ray (AB || CD), but laterally displaced.

• LATERAL DISPLACEMENT is represented as DE in the diagram.

Hope It Helps.

Answer 2

Explanation:

LATERAL SHIFT IN REFRACTION THROUGH SLAB:

\setlength{\unitlength}{1cm}\begin{picture}(12,12)\put(0,0){\framebox(9,5)}\multiput(4,6.5)(2,-5){2}{\line(0,-1){3}}\thicklines\qbezier(4,5)(4,5)(6,0)\qbezier(1.5,7.5)(3,6)(4,5)\qbezier(6,0)(6,0)(8.5,-2.5)\multiput(4,5)(0.5,-0.5){14}{\line(1,-1){0.4}}\linethickness{0.24mm}\qbezier(3.5,5.5)(3.7,5.9)(4,5.7)\qbezier(6,-0.6)(6.3,-0.85)(6.5,-0.5)\qbezier(4,4)(4.3,3.8)(4.38,4.1)\qbezier(6,1)(5.8,1.2)(5.62,0.9)\put(2.5,6.5){\line(-1,0){0.4}}\put(2.5,6.5){\line(0,1){0.4}}\put(7.5,-1.5){\line(-1,0){0.4}}\put(7.5,-1.5){\line(0,1){0.4}}\qbezier(5,2.5)(5,2.5)(4.6,2.7)\qbezier(5,2.5)(5,2.5)(5.2,2.9)\thinlines\qbezier(8,-2)(8,-2)(9.5,-0.5)\put(3.8,6.7){\bf M}\put(3.8,3){\bf M'}\put(5.8,1.8){\bf N}\put(5.8,-2){\bf N'}\put(1,7.7){\bf A}\put(4.3,5.3){\bf B}\put(5.4,-0.5){\bf C}\put(7.5,-2.4){\bf D}\put(9.8,-0.5){\rm E}\put(0.5,5.5){\sf AIR}\put(0.5,2.5){\sf GLASS}\put(0.5,-0.8){\sf AIR}\put(3.5,6){\bf i}\put(6.4,-1.2){\bf e}\linethickness{0.24mm}\qbezier(4,3.85)(4.3,3.7)(4.4,3.9)\qbezier(6,1.15)(5.8,1.3)(5.6,1.1)\put(5,5.5){\sf DE=Lateral Displacement}\end{picture}

Here, you can see that a light ray incident on a rectangular slab is undergoing refraction twice at the two parallel surfaces.

At the 1st interface, the ray refracts towards the normal.

At the 2nd interface, the ray refracts away from the normal.

This results in the final emergent ray to be parallel to the incident ray (AB || CD), but laterally displaced.

LATERAL DISPLACEMENT is represented as DE in the diagram.

Hope It Helps.