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

`m = 4//3`

Explanation

Applying `(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2)-mu_(1))/(R)`
First on plane surface
`(1.5)/(v_(1)) - (1)/((-mR)) = (1.5-1)/(oo) (R = oo)`
`:. V_(1) = (-1.5 mR)`
Then on curved surface
`(1)/(oo) - (1.5)/(-(1.5mR + R)) = (1-1.5)/(-R)` [`v = oo` becuase final image is at infinity]
`rArr (1.5)/((1.5m + 1)R) = (0.5)/(R) rArr 3 = 1.5m +1`
`rArr (3)/(2)m = 2` or `m = 4//3`