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Two thin sysmmetrical lenses of differen...

Two thin sysmmetrical lenses of different nature and of different material have equal raii of curvature `R=15cm`. The lenses are put close together and immersed in water `(mu_(w)=4//3)` . The focal length of the system ini water is 30cm. The difference between refractive indices of tthe two lenses is

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Let `f_(1)` and `f_(2)` be the focal length in water. Then
`1/(f_(1))=(mu_(l)/mu_(w) -1)(1/R+ 1/R)implies 1/(f_(1))=(mu_(.)/mu_(w)-1)(2/R)to (1)`
`1/(f_(2))=(mu_(2)/mu_(w)-1)(- 2/R)to (2)`
Adding Eqs. (1) and (2) we get
`1/(f_(1)) + 1/f_(2))=(2(mu_(1)-mu_(2)))/(mu_(w)R)`
But the given system is equal to combination of two lens kept in contact in liquid so
`1/F =1/(F_(1)) + 1/(f_(2))` or `1/30 = (2(mu_(1)-mu_(2)))/(mu_(w)R)`
`:. (mu_(1)-mu_(2))=(4xx1)/(3xx60)=1/3`
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