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.

Options

A. `3//2`
B. `4//3`
C. `5//4`
D. None of these

Answer

B

Explanation

Apply `(mu_(2))/(v)-(mu_(1))/(u)+(mu_(2)-mu_(1))/(R), (mu_(2))/(v)=(mu_(1))/(u)` ltBrgt (as R = `oo`)
`therefore` Image distance v = `(mu_(1))/(mu_(2))u=(1.5)/(1.0)(-mR)`
=-1.5mR
For curved surface
Here, `(1)/(oo)+(1.5)/((1.5m+1)R)=(1-1.5)/(-R)" "thereforem=(4)/(3)`