Refraction at Plane and Spherical surfaces

REFRACTION AT A PLANE SURFACE

Laws of Refraction (Snell's Law)

(i) The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

(ii) The ratio of the sines of the angle of incidence (i) and of the angle of refraction (r) is a constant quantity $\mu$ for two given media, which is called the refractive index of the second medium with respect to the first.

$\frac{\sin i}{\sin r}=cons\tan t=\mu$

When light propagates through a series of layers of different medium as shown in the figure, then the Snell's law may be written as

$\mu$₁ sin$\theta$₁ = $\mu$₂ sin$\theta$₂ = $\mu$₃ sin$\theta$₃ = $\mu$₄ sin$\theta$₄ = constant

In general, $\mu$ sin$\theta$ = constant

When light passes from rarer to denser medium it bends toward the normal as shown in the fig.

According to Snell's law

$\mu$₁ sin$\theta$₁ = $\mu$₂ sin$\theta$₂

When a light ray passes from denser to rarer medium it bends away from the normal as shown in the fig.

Fig. A light ray passing from water to air bends away from the normal.

For a given point object, the image formed by refraction at plane surface is illustrated by the following diagrams.

Determination of the image formed for the cases given becomes much simpler when we restrict ourselves to nearly normal incident rays.

CASE – I (Object in the denser medium)

For nearly normally incident rays$\theta$₁ and$\theta$₂ will be very small.

$\tan {\theta }_1=\sin {\theta }_1=\frac{AB}{Objectdis\tan cefromtherefractingsurface}$

Similarly,

$\sin {\theta }_2=\frac{AB}{\mathrm{Im}agedis\tan cefromtherefractingsurface}$

$\Rightarrow \frac{\mathrm{Im}agesdistancefromtherefractingsurface}{Objectdis\tan cefromtherefractingsurface}$

$=\frac{\sin {\theta }_1}{\sin {\theta }_2}{=}_1{\mu }_2=\frac{{\mu }_2}{{\mu }_1}$

The same result is obtained for the other case also. The image distance from the refracting surface is also known as Apparent depth or height.

Apparent Shift

Apparent shift = Object distance from refracting surface – image distance from refracting surface.

$\Delta$y (apparent shift) $t=\left(1-\frac{1}{\mu }\right)$ where t is the object distance and $\mu =\frac{{\mu }_1}{{\mu }_2}$

$\bullet$ If there are a number of slabs with different refractive indices placed between the observer and the object.

Total apparent shift = $\sum \Delta$y_i

Illustration 1: A person looking through a telescope T just sees the point A on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid ($\mu$ = 1.5), he observes a mark at the centre, B, of the vessel. What is the height of the vessel if the diameter of its cross-section is 10cm?

Solution: It is mentioned in the problem that on filling the vessel with the liquid, point B is observed for the same setting; this means that the images of point B, is observed at A, because of refraction of the ray at C.

For refraction at C,

$\frac{\sin r}{\sin i}{=}_a{\mu }_l=1.5$

Now sin$r=\frac{AD}{AC}=\frac{10}{\sqrt{{10}^2+h^2}}$

Where h is the height of vessel.

\begin{align}  \sin \,i=\dfrac{BD}{BC}=\dfrac{5}{\sqrt{{{5}^{2}}+{{h}^{2}}}} \\  \therefore \dfrac{10}{\sqrt{100+{{h}^{2}}}}.\dfrac{\sqrt{25+{{h}^{2}}}}{5}=1.5 \\  \therefore \dfrac{25+{{h}^{2}}}{100+{{h}^{2}}}=\dfrac{9}{16} \\  \therefore 100\,+\,16{{h}^{2}}=900+9{{h}^{2}} \\  \therefore 7{{h}^{2}}=500\therefore \,h=8.45\,cm \\ \end{align}

REFRACTION AT SPHERICAL SURFACE

When light strikes a spherical (curved) surface, the position of object (O), the image (I) and the radius of curvature (R) are related as:

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$; here $\mu$₂ > $\mu$₁

Consider another case, a spherical surface of radius R separating two media with refractive indices m₁ & m₂. The radius of curative R will be taken positive when incident rays strike the convex side of the surface. If u is the object distance and v is the image distance,

For light rays going from medium 1 ($\mu$₁) to medium 2 ($\mu$₂):

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

The corresponding equation for refraction at a plane surface will be:

$\frac{{\mu }_1}{u}+\frac{{\mu }_2}{v}=0$ (taking R = $\infty$)