A quarter cylinder of radius R and refractive index 1.5 is placed on a table.A point object P is kept at a distance of mR from it. Find the value of m for whicha ray from P will emerge parallel to the table as shown in the figure.

Answer

C, D

Explanation

First of all, we consider the refraction the refraction at plane surface. Here the image of P will from `I^(`)` after refraction from I surface.
For plane surface:
Object distance`u=-mR`
Radius of curvature of the plane surface`=infty`
The ray is coming from air and incident on the glass.
Here `mu_(1)=1`,`mu_(2)=1.5`
Apply `mu_(2)/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)`,`(mu_(2))/(v)=(mu_(1))/(u)`... `(as R=infty)`
`:'` Image distance `v=(mu_(1))/(mu_(2))u=(1.5)/(1.0)(-mR)=-1.5mR`
Now we consider refraction at the curved surface. Object distance, `u=-(1.5mR+R)`
Here, `mu_(2)=1`,mu_(1)=1.5`, Image distance, `v=infty`,
Radius of curvature`=-R`
Here, `(1)/(infty)+(1.5)/((1.5m+1)R)=(1-1.5)/(-R)` `:'` `m=(4)/(3)`