Two thin symmetric glass lenses, one biconvex and the other biconcave, are brought into contact to make a system with optical power `Phi = 0.50 D`. Newton's rings are observed in reflected light with wavelength `lambda = 0.61 mu m`. Determine:
(a) the radius of the tenth dark ring,
(b) how the radius of that ring will change when the space between the lenses is filled up with water.

Explanation

Here `Phi = (n - 1) ((2)/(R_(1)) - (2)/(R_(2)))`
so `(1)/(R_(1)) - (1)/(R_(2)) = (Phi)/(2(n - 1))`.
As in the previous example, for the dark rings we have
`r_(k)^(2) ((2)/(R_(1)) - (2)/(R_(2))) = (Phi)/(2(n - 1)) r_(k)^(2) = k lambda`
`k = 0` is dark spot, excluding it, we take `k = 10` hre.
Then `r = sqrt(20 lambda(n -1))/(Phi) = 3.49 mm`.
(b) Path difference in water film will be
`n_(0)bar(r^(2)) ((2)/(R_(1)) - (2)/(R_(2)))`
where `bar(r) =` nwe radius of the ring. Thus
`n_(0)bar(r^(2)) = r^(2)`
or `bar(r) = r// sqrt(n_(0)) = 3.03 mm`.
Where `n_(0) = R.I`. of water `= 1.33`.