A ray of light passes through a transparent sphere of refractive index `mu` and radius R. If b is the distance between the incident ray and a parallel diameter of

Explanation

Following is the ray diagram for the given situation

In the above figure `i` is the angle of incident at the first point P and then light ray is refracted at an angle r with the normal . Using geometry of circle we can easily understand that angle of incidence at the point Q will also be equal to r and hence in air light ray will be refracted at an angle i with the normal . Hence in both the refractions light ray bends by an amount i- r in the same sense which is clockwise in the figure . If `theta` is the angle of deviation then we can write the following :
`theta = 2 (i - r)`
`implies " " (theta)/(2) = i - r `
`implies " " sin ""(theta)/(2) = sin (i- r)`
`implies " " sin ""(theta)/(2) = sin i cos r - cos i sin r " " ... (i)`
From the figure we can write the following :
`sin i = (b)/(R) " " ... (ii)`
`cos i = sqrt(1 - sin^(2) i)`
`implies " " cos i = sqrt( 1 - (b^(2))/(R^(2)))`
`implies cos i = (1)/(R) sqrt( R^(2) - b^(2)) " " .... (iii)`
Using Snell.s law we can write the following
`mu sin phi =` constant
`implies " " 1 xx sin i = mu xx sin r`
`implies " " sin r = (b)(mu R) ... (iv)`
`implies " " cos r = sqrt(1 - sin^(2) r)`
`implies " " cos r = sqrt(1 - (b^(2))/(mu^(2) R^(2)))`
`implies cos r = (1)/(mu R) sqrt(mu^(2) R^(2) - b^(2)) " " ... (v)`
Now substituting values from equation (ii) , (iii) , (iv) and (v) in equation (i) we get the following :
`sin"" (theta)/(2) = sin i cos r - cos i sin r `
`implies sin ""(theta)/(2) = (b)/(R) xx (1)/(muR) sqrt(mu^(2) R^(2) - b^(2)) - (1)/(R) sqrt(R^(2) - b^(2)) xx (b)/(muR)`
`implies sin ""(theta)/(2) = (b)/(muR^(2)) [ sqrt(mu^(2) R^(3) - b^(2)) - sqrt(R^(2) - b^(2))]`
Hence we have proved the desired relation .