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There are two thin symmetrical lenses : ...

There are two thin symmetrical lenses : one is converging , with refractive index `n_(1) = 1.70`, and the other is diverging with refractive index `n_(2) = 1.51`. Both lenses have the same curavature close together and submerged into whater. What is the focal length of this system in water ?

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Focal length of the converging lens, when it is submerged in water of `R.I. n_(0)` (say):
`(1)/(f_(1)) = ((n_(1))/(n_(0)) - 1) ((1)/(R ) - (1)/(R )), (2(n_(1) - n_(0)))/(n_(0)R)` (1)
Similarly, the focal length of divierging lens in water.
`(1)/(f_(2)) = ((n_(1))/(n_(0)) - 1) ((1)/(-R ) - (1)/(R )), (-2(n_(1) - n_(0)))/(n_(0)R)` (2)
Now, when they are put together in the water, the focal length of the system,
`(1)/(f) = (1)/(f_(1)) + (1)/(f_(2))`
`= (2(n_(1) - n_(2)))/(n_(0)R) - (2(n_(2) - n_(0)))/(n_(0)R) = (2(n_(1) - n_(2)))/(n_(0)R)`
or, `f = (-n_(0)R)/(2(n_(1) - n_(2))) = 35 cm`
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