Answer:
Option D: 1.5in in front of the target
Explanation:
The object distance is [tex]y= 6in[/tex].
Because the surface is flat, the radius of curvature is infinity .
The incident index is [tex]n_i=\frac{4}{3}[/tex] and the transmitted index is [tex]n_t= 1[/tex].
The single interface equation is [tex]\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}[/tex]
Substituting the quantities given in the problem,
[tex]\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0[/tex]
The image distance is then [tex]y^i=-\frac{18}{4}in =-4.5in[/tex]
Therefore, the coin falls [tex]1.5in[/tex] in front of the target