A monochromatic beam of light is incident at `60^@` on one face of an equilateral prism of refractive inder `n` and emerges from the opposite face making an angle `theta` with the normal. For `n = sqrt(3)`, the value of `theta` is `60^@ and (d theta)/(dn) = m`. The value of `m` is.

Answer

2

Explanation

Applying Snell.s law
`rArr 1 xx sin 60 = n xx sin r_(1)`
`rArr sin r_(1) = (sqrt(3))/(2n)`
`rArr cos r_(1) = (1)/(2n) sqrt(4n^(2) - 3)` ….(1)
`mu sin theta = const`
`rArr n xx sin r_(2) = 1 xx sin theta`
`rArr sin theta = n sin r_(2)`
`rArr sin theta = n sin (50 - r_(1))`
`rArr sin theta = n[sin 60 cos r_(1) - cos 60 sin r_(1)]`
`rArr sin theta = n[(sqrt(3))/(2) cos r_(1) - (1)/(2) sin r_(1)]`
`rArr sin theta = n xx [(sqrt(3))/(2) xx (1)/(2n)sqrt(4n^(2) - 3) - n xx (1)/(2) xx (sqrt(3))/(2n)]`
`rArr sin theta (sqrt(3))/(4)sqrt(4n^(2) - 3) - (sqrt(3))/(4)`
On differentiating both the sides we get the following :
`rArr cos theta xx (d theta)/(d n) = (sqrt(3))/(4) xx (8n)/(2sqrt(4n^(2)) - 3)`
Now substituting `n = sqrt(3)` and `theta = 60^(@)`.
`rArr cos 60 xx (d theta)/(dn) = (sqrt(3))/(4) xx (8sqrt(3))/(6)`
`rArr (1)/(2) xx (d theta)/(dn) = (sqrt(3))/(4) xx (8sqrt(3))/(6)`
`rArr (d theta)/(dn) = 2`
Hence answer is 2.