A) \[\left[ 1-\frac{1}{\mu t} \right]\]
B) \[\left[ t-\frac{\mu }{t} \right]\]
C) \[t\left[ 1-\frac{t}{\mu } \right]\]
D) \[[\mu t-1]\]
Correct Answer: C
Solution :
The situation is shown in figure. Because of the refraction at the first surface, the image of \[O\] is formed at \[{{O}_{1}}\]. For this refraction, the real depth is \[AO=x\] and the apparent depth is \[A{{O}_{1}}\]. Also, the first medium is air and the second is the slab. Thus,\[\frac{x}{A{{O}_{1}}}=\frac{1}{\mu }\,\] or \[A{{O}_{1}}=\mu x\] The point Oi acts as the object for the refraction at the second surface. Due to this refraction, the image of \[{{O}_{1}}\] is formed at \[I\]. Thus, \[\frac{B{{O}_{1}}}{BI}=\mu \] or \[\frac{AB+B{{O}_{1}}}{BI}=\mu \] or \[\frac{t+\mu x}{BI}=\mu \] or \[BI=x+\frac{t}{\mu }\] The net shift is \[OI=OB=\,(x+t)-\,\left( x-\frac{t}{\mu } \right)\] \[=t\,\left( 1-\frac{1}{\mu } \right)\] Which is independent of \[x\].You need to login to perform this action.
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