Refraction Of Light

Refraction Of Light
 

REAL AND APPARENT DEPTHS(NORMAL SHIFT) 
    As a consequence of refraction, the depth of an object lying inside an optically denser medium appears to be less than its real depth.

Obect is in denser madium Observer is in rarer medium:

    Suppose that the surface XY separates medium a (air) and medium b (water or any other transparent liquid). Let O be a point object lying inside the liquid. A ray of light OA incident along normal to the surface of the liquid, will be refracted as such along AL. Another ray of light OB, on refraction into air, bends away from the normal NBN’ and travels along BC. The two refracted rays appear to come from the point I. Therefore, the object, though lying at depth AO, appears to be at depth AI. Accordingly, AO is called the real depth and AI, the apparent depth of the object O. 

    Suppose that \( \angle OBN' = i \) and \( \angle NBC = r \). Then, refractive index of medium a (air) w.r.t medium b (water) is given by
        \( ^b \mu _a = \frac{{\sin i}} {{\sin r}} \)
    since \( ^a \mu _b = \frac{1} {{^b \mu _a }} \), we have \( ^a \mu _b = \frac{{\sin i}} {{\sin r}} \)    
    Now, \( \angle AOB = \angle OBN' = i \)
Therefore, from the right angled \( \Delta OAB \), we have \( \sin i = \frac{{AB}} {{OB}} \)

Also, \( \angle AIB = \angle NBC = r \)
    Therefore, from the right angled \( \Delta IAB \), we have \( \sin r = \frac{{AB}} {{IB}} \) 
    Substituting for sin i and sin r in the equation, we have
      \( ^a \mu _b = \frac{{AB/IB}} {{AB/OB}} = \frac{{OB}} {{IB}} \)  
    Since the aperture of the eye is very small, the two rays AL and BC will enter the eye only, if point B lies very close to the point A. In such a situation,\( OB \approx OA \) , the real depth of the object and \( IB \approx IA \), the apparent depth of the object
    \( ^a \mu _b = \frac{{OA}} {{IA}} = \frac{{real\text{ }depth}} {{apparent\text{ }depth}} \)
    The object appears to be raised from its real position 0 to its apparent position I. The distance OI through which the position of the object appears to be raised, is called normal shift and its denoted by d. Therefore, normal shift, d = OA - IA
                \( = OA\left( {1 - \frac{{IA}} {{OA}}} \right) = OA\left( {1 - \frac{1} {{OA/IA}}} \right) \)
             or \( d = t\left( {1 - \frac{1} {{^a \mu _b }}} \right) \)
It may be pointed out that the normal shift in the position of the object depends upon
i)    the real depth of the object i.e., the thickness of the refracting medium and
ii)    the refractive index of the refracting medium reciprocal of the refractive index of the medium a w.r.t the medium b.

Object is in rarer medium Observer is in denser medium:
    When the object in rarer medium (air) at a distance ‘y’ from boundary and an observer near to normal in denser medium of refractive index \( '\mu ' \). By ray diagram in figure it is observed that the image is virtual, on same side to boundary and its distance from the boundary is \( '\mu ' \) times the object distance. Since \( \mu > 1 \) image distance is more than object distance.

\( \sin i \approx \tan i = \frac{{AB}} {{AO}},\sin r \approx \tan r = \frac{{AB}} {{AI}} \)
    According to Snell’s law \( 1.\sin i = \mu \sin r \)
    \( \frac{{AB}} {{AO}} = \mu \frac{{AB}} {{AI}},AI = \mu .AO \)
    Therefore apparent height of object \( \left( {AI} \right) = \mu \times \text{real height of object}\left( {AO} \right) \)
    i. e.,\( y_{app} = \mu .y_{real} \) 
    Apparent shift = AI - AO
    Apparent shift \( = \left( {\mu - 1} \right)y \)