4.How does the lateral shift change with the increase in the angle of incidence?

The lateral shift is the perpendicular distance between the path of the ray of light when it emerges from the refracting medium and the path that the ray would have taken had it not been refracted.

⇒$\Rightarrow$ The lateral shift is CF,$CF,$ where CF⊥BE.$CF\perp BE.$

The angle of incidence is i$i$ and the angle of refraction is r.$r.$

⇒CF=BCsin(∠CBF)=BCsin(i−r).$\Rightarrow CF=BC\sin (\mathrm{\angle }CBF)=BC\sin (i-r).$

Let the thickness of the glass slab, BG$BG$ be t.$t.$

⇒BC=BGcosr=tcosr.$\Rightarrow BC=\frac{BG}{\cos r}=\frac{t}{\cos r}.$

⇒CF=(tcosr)sin(i−r)=(tcosr)(sinicosr−cosisinr)$\Rightarrow CF=\left(\frac{t}{\cos r}\right)\sin (i-r)=\left(\frac{t}{\cos r}\right)(\sin i\cos r-\cos i\sin r)$

=t(sini−cositanr).$=t(\sin i-\cos i\tan r).$

sinisinr=μ,$\frac{\sin i}{\sin r}=\mu ,$ the refractive index.

⇒sinr=siniμ.$\Rightarrow \sin r=\frac{\sin i}{\mu }.$

⇒tanr=siniμ2−sin2i√.$\Rightarrow \tan r=\frac{\sin i}{\sqrt{{\mu }^2-{\sin }^{2}i}}.$

⇒CF=t(sini−cosisiniμ2−sin2i√)=t(sini−sin2i2μ2−sin2i√).$\Rightarrow CF=t(\sin i-\frac{\cos i\sin i}{\sqrt{{\mu }^2-{\sin }^{2}i}})=t(\sin i-\frac{\sin 2i}{2\sqrt{{\mu }^2-{\sin }^{2}i}}).$

⇒$\Rightarrow$ The variance of the lateral shift with the angle of incidence is given by the formula,

$$ Lateral shift =t(sini−sin2i2μ2−sin2i√).www.totallearn.blogspot.com