A cylinderical glass rod of radius `0.1 m` and refractive index `sqrt3` lies on a horizontal plane mirror. A horizontal ray of light moving perpendicular to the axis of the rod is incident on it. At what height from the mirror should the ray be incident so that it leaves the rod at a height of `0.1 m` above the plane mirror? At what centre to centre distance a second similar rod, parallel to the first, be placed on the mirror, such that the emergent ray from the second rod is in line with the incident ray on the first rod?

Answer

A, C

Explanation

(i) `PO = OQ`
`:. angleOPQ = angleOQP = r("say")`

Also, `i = r + r = 2r`
In `DeltaPQR, h = OP sin I = 0.1 sin t`
`= 0.1 sin 2r`
or `h = .0.2 sim r cosr`……(i)
Also `sqrt(3) = (sin i)/(sin r) = (2 sin r cos r)/(sin r)`
`= 2 cos r `
`:. r = 30^(@)`
Substituting in Eq. `(i)`, we get
` h = 0.2 xx (1)/(2) xx (sqrt(3))/(2)`
`= 0.086 m`.
Hence, height from the mirror is
`0.1 + 0.086 = 0.186 m`
(ii) Use the principle of reversiblity

`i = 2r = 60^(@)`
Now, `(QS)/(MS) = coti = cot 60^(@) = (1)/(sqrt(3))`
`:. QS = (MS)/(sqrt(3)) = (0.1)/(sqrt(3))`
`:.` The disered distance,
`OC = 2 xx 0.1 + (2 xx 0.1)/(sqrt(3))`
`0.315`