The critical angle is given by sinC = 1/n.
The parallel beam of light making more than critical angle with the curved surface of the quarter cylinder, will not come out of the cylinder. The ray making the angle of incidence equal to critical angle will come out of the glass tangential to the cylindrical surface. If O is the center and PQ is the tangent, in the right angled triangle OPQ angle POQ is equal to C and cosC will give you the first image of the ray.
When the ray is very close to the base of the cylinder the image position is given by
n/u + 1/v = (n1)/R. Here u = infinity, R = 5 cm. If S is the image of this ray, QS is the image on the table due to the parallel beam of light.
