Twot thin lenses when placed in contact, then the power of combination is +10D. If they are kept 0.25m apart, then the power reduceds to +6D. The focal lengths of the lenses (in m) will be

Explanation

Let `P_(1)` and `P_(2)` be the powers of the two thin lenses, repectively. Power of the two lenses in contact`=P_(1)+P_(2)`. Power of the two lenses at a distance `x=P_(1)+P_(2)-xP_(1)P_(2)`. From the given data,
we get
`P_(1)+P_(2)=10m^(-1)` and `P_(1)+P_(2)-(0.250)P_(1)P_(2)=6m^(-1)`
Form these two expression, we get
`P_(1)+P_(2)16m^(-2)` and `P_(1)-P_(2)=sqrt((P_(1)+P_(2))^(2)-4P_(1)P_(2))`
`=sqrt((10^(-1))^(2)-4(16^(-1)))=6m^(-1)`
Thus, `P_(1)+P_(2)=10m^(-1)` and `P_(1)+P_(2)=6m^(-1)` , we get
`P_(1)=8m^(-1)and P_(2)=2m^(-1)`
Hence , `f_(1)=(1)/(P_(1))=(1)/(8)m=0.125m`
and `f_(2)=(2)/(P_(2))=(1)/(2)m=0.5m`