A rectangular glass block of thickness 10 cm and refractive index 1.5 placed over a small coin. A beaker is filled with water of refractive index `4//3` to a height of 10 cm and is placed over the glass block.
(a) Find the apparent position of the object when it is viewed at near normal incidence.
(b) if the eye is slowly moves away from the normal at a certain position, the object is found to disappear, due to total internal reflection. At what surface does this happen and why ?

Explanation

Given,

Critical angle,
`mu_(2)sinC=mu_(1)sin90^(@)`
`2sinC=sqrt2rArrsinC=1/sqrt(2)=sin45^(@)`
`C=45^(@)`
For T.I.R at AB, i`gt`C i.e., i`gt45^(@)`

criticle angle
`mu_(2)sinC=mu_(3)sin90^(@)`
`2sinC=sqrt3rArrsinC=sqrt3/2=sin60^(@)`
`C=60^(@)`
For T.I.R at CD, i`gt`C o.e., i`gt60^(@)`
`i_(min)=60^(@)`

(a) Total shift,
`OI=t_(1)(1-1//mu_(1))+t_(2)(1-1//mu_(2))`
`=10(1-1/(3//2))+10(1-1/(4//3))`
`=10/3+10/4`
Distance of image from P,
`PI=PO-Ol=(10+10)-(10/3+10/4)`
=14.17cm
(b) The image of coin disappears when the ray suffers TIR at either at AB or CD
The critical angles for glass-water surface
`mu_(1)sinC_(1)=mu_(2)sin90^(@)`
`3/2sinC_(1)=2/3`
`sinC_(1)=4/9`
`C_(1)sin^(-1)(4//9)`
The critical angles for water-air surface
`mu_(2)sinC_(2)=1xxsin 90^(@)`
`2/3sinC_(2)=1`
`sinC_(2)=3/2`
`C_(2)ltC_(1)` TIR occurs earlier at the water-air surface as eye is moved away from the normal.