Question

Derive an expression for lateral shift and normal shift. On what factors these depends

Answer 1

$Lateralshift:itisdefineastheperpendiculardistancebetweenincidencerayandemergeneceray,$

$aslighttravelthroughrectengularglassslab.$

$Lateralshift;\Delta =CE$

$\Delta =BCsin(i-r)$

$\Delta =\frac{BN}{cosr}sin(i-r)$

$\Delta =\frac{tsin(i-r)}{cosr}$

$Thisaboveequationisknownaslateraldisplacement\left(shift\right)forsinglerectangularglassslab.$

$Tocalculatelateralshiftbymultipleslab.$

$\Delta =\frac{t_{1}sin(i-r_1)}{cos{r}_1}+\frac{t_{2}sin(i-r_2)}{cos{r}_2}+\frac{t_{3}sin(i-r_3)}{cos{r}_3}+...\frac{t_{n}sin(i-r_n)}{cos{r}_n}$

$\therefore ThereforeLateralshiftdependsuponfollowingparameters:$

$\left(1\right)thicknessoftheglassslab$

$\left(2\right)angleofincidence$

$\left(3\right)angleofrefraction$

$Normalshift:itistheapparentshiftinthepositionofanobjectplacedinonemediumandviewedalongthenormal,fromtheothermedium.$

$case\left(1\right)whenobjectisplacedindensermedium$

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=0$

$\frac{{\mu }_2}{v}=\frac{{\mu }_1}{u}$

$v=u\left(\frac{{\mu }_2}{{\mu }_1}\right);\because {\mu }_1>{\mu }_2,\therefore {\mu }_1=\mu \&{\mu }_2=1$

$apparentdistance=actualdistance\left(\frac{1}{\mu }\right)$

$\therefore shiftofobject=actualdistance-apparentdistance$

$shiftofobject=actualdistance(1-\frac{1}{\mu })$

similarly, for object is in rarer medium also,

$shiftinobject=actualdistance(\mu -1)$

Therefore we can say, normal shift depends upon following parameters :

(1) actual distance

(2) refractive index of the medium