Subject: Physics
An object placed in a denser medium, when viewed from rarer medium appears to be at a lesser depth than its real depth due to refraction of light.
Consider a point object O at the bottom of a beaker containing water. Suppose XY is the plane surface which separated air and water. A ray OA from O incident normally to the surface XY passes without bending along AD. Another ray OB is refracted away from normal along BC. When viewed from above, the rays will appear to come from point I which is the point of intersection of OD and BC produced backward. Thus, a virtual image of O is formed at I. Therefore, new depth the object is AL. The depth AO is the real depth of the object, and depth AI is called apparent depth which is as shown in the figure.
\begin{align*} \text {From Snell’s law} \\ _a\mu _w &= \frac {\sin r}{\sin i} \\ \text {where I is angle of incident and r is the angle of refraction. } \\ \text {In} \Delta ABO, \: \sin I &= \frac {AB}{OB} \\\text {In} \Delta ABI, \: \sin r &= \frac {AB}{IB} \\ \text {Then,} \\ _a\mu _w &= \frac {\sin i}{\sin r} \\ &= \frac {AB/IB}{AB/OB} \\ \text {If point B is very close to A, then we have } \: OB = OA \text {and} \: IB = IA. \text {So,} \\ _a\mu _w &= \frac {OA}{IA} = \frac {\text {real depth}} {\text {Apparent depth}} \dots (i) \end{align*}
Apparent shift
If the real depth, OA = t, then from equation (i), apparent depth \( = \frac {1}{_a\mu _w} \). The apparent shift of the object is given by
\begin{align*} d &= OI = OA-AI \\ &= \text {real depth} - \text {apparent depth} \\ &= \frac {1}{_a\mu _w} \\ \text {or,} &= t \left ( 1-\frac {1}{_a\mu _w} \right ) \\ \end{align*}
When a ray of light passes from a denser medium to rarer medium, it bends away from the normal. Thus, an incident ray AO passes from water to air and bends away from normal in the direction OB. As the angle of incidence is increased, the angle of refraction also increases and becomes 90^{o} at a particular angle of incidence. The angle of incidence in denser medium for which the angle of refraction in rarer medium is 90^{o}, is called the critical angle C of the medium.
\begin{align*} \text {At this condition,} \\ _w\mu _a &= \frac {\sin i}{\sin r} \\ &= \frac {\sin c}{\sin 90^o} \\ &= \sin c \\ \text {or,} \: _a\mu _w &= \frac {1}{\sin c} \\ \therefore c &= \sin ^{-1} \left (\frac {1}{_a\mu _w}\right ) \\ \end{align*}
If the angle of incidence in the denser medium is less than the critical angle, the light is refracted in the normal way. But is the angle is greater than the critical angle, the light is totally internally reflected in the same denser medium? This is known as total internal reflection. In the figure, AH represents a ray of light whose angle of independence (ÐAHN) is greater than critical angle c and HK is the reflected ray in the same medium which obeys the laws of reflection of light.
Conditions for Total Internal Reflection
Applications
Conditions for Total Internal Reflection
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