Refraction From A Plane Surface And T.I.R Spherical Surface (JEE LEVEL)

Consider a point object O placed in front of a spherical surface separating two mediums. The incident rays OD and OP  meet at point I after refraction. The line NC represent a normal at the point of incidence D. Here C be the center of curvature of the surface.

 From the figure      

  i=\alpha +\beta                     …(i)          (being exterior angle of \displaystyle \Delta OCD)

Also

\beta =r+\gamma                                    (being exterior angle of \displaystyle \Delta CDI)

 or   r=\beta -\gamma                   …(ii)

Also by Snell’s law

\frac{{{{\mu }_{2}}}}{{{{\mu }_{1}}}}=\frac{{\sin i}}{{\sin r}}=\frac{i}{r}      (if i & r are small) 

hence from (i) and (ii)

\frac{{{{\mu }_{2}}}}{{{{\mu }_{1}}}}=\frac{{\alpha +\beta }}{{\beta -\gamma }}, 

By cross-multiplication,

{{\mu }_{1}}(\alpha +\beta )={{\mu }_{2}}(\beta -\gamma )              

\Rightarrow      {{\mu }_{1}}\alpha +{{\mu }_{1}}\beta ={{\mu }_{2}}\beta -{{\mu }_{2}}\gamma            

\Rightarrow     {{\mu }_{1}}\alpha +{{\mu }_{2}}\gamma =({{\mu }_{2}}-{{\mu }_{1}})\beta

because if \displaystyle \alpha ,\beta ,\gamma are very small then \tan \alpha =\alpha ,\tan \beta =\beta ,\tan \gamma =\gamma hence we can write the above equation as

{{\mu }_{1}}\tan \alpha +{{\mu }_{2}}\tan \gamma =({{\mu }_{2}}-{{\mu }_{1}})\tan \beta  

{{\mu }_{1}}\frac{h}{{\mathbf{QO}}}+{{\mu }_{2}}\frac{h}{{\mathbf{QI}}}=({{\mu }_{2}}-{{\mu }_{1}})\frac{h}{{\mathbf{QC}}}

If \displaystyle \alpha is small i.e., rays are paraxial & R is large, then PQ is a negligible distance hence point Q will be very close to point P therefor we can replace point Q by P so we get

{{\mu }_{1}}\frac{1}{{\mathbf{PO}}}+{{\mu }_{2}}\frac{1}{{\mathbf{PI}}}=({{\mu }_{2}}-{{\mu }_{1}})\frac{1}{{\mathbf{PC}}}

Now from the figure put OP = -u , PI = v and PC = R to get the final formula

\frac{{{{\mu }_{1}}}}{{-u}}+\frac{{{{\mu }_{2}}}}{v}=\frac{{{{\mu }_{2}}-{{\mu }_{1}}}}{R}

or    \displaystyle {\frac{{{{\mu }_{2}}}}{v}-\frac{{{{\mu }_{1}}}}{u}=\frac{{{{\mu }_{2}}-{{\mu }_{1}}}}{R}}

Note:

If first medium is air, then \displaystyle {{\mu }_{1}} becomes 1, hence the formula becomes \frac{\mu }{v}-\frac{1}{u}=\frac{{\mu -1}}{R}.

Important points:

(i) 

These equations are valid for all refracting surfaces convex, concave or plane. In case of plane refracting surface the radius R can be taken as   \infty .

(ii)

The rules for signs for a single refracting surface are same as for spherical mirrors.

Illustration

A ray of light is incident at an angle of \displaystyle {{60}^{o}} on one face of a rectangular glass slab of thickness 0.1 m and refractive index 1.5. Calculate the lateral shift produced.

Solution

Here    i = \displaystyle {{60}^{o}}; \displaystyle \mu = 1.5 and t = 0.1 m

Now,   \displaystyle \mu = \frac{{\sin \,i}}{{\sin \,\,r}} 

or    sin r = \frac{{\sin \,i}}{\mu }\,=\,\,\frac{{\sin \,\,60{}^\circ }}{{1.5}} = \frac{{.866}}{{1.5}} = .5773

or   r = \displaystyle {{35}^{o}}{{6}^{\prime}}

Now, lateral shift       

d = \frac{t}{{\cos \,r}}\,\,\sin \,\,(i\,-\,r) 

   =   \frac{{0.1}}{{\cos \,35{}^\circ {{6}^{\prime}} }} sin ( \displaystyle {{60}^{o}}-{{35}^{o}}{{6}^{\prime}})

   = 0.0512 m

Illustration

A person looking through the telescope just sees a point on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid of refractive index 1.5, he observes a mark at the centre B of the bottom, without moving the telescope or the vessel. What is the height of the vessel, if the diameter of its cross-section is 10cm?

Solution

Consider a ray from B along the normal and the other as shown in figure. Let h be the height of the vessel. Then 

\sin \theta =\frac{5}{{\sqrt{{{{h}^{2}}+{{5}^{2}}}}}} 

and \sin {{\theta}^{\prime}} =\frac{5}{{\sqrt{{\frac{{{{h}^{2}}}}{4}+{{5}^{2}}}}}}