A thin convex lens which is made of glass (refractive index 1.5) has a focal length of 20 cm.it is now completely immersed in a transparent liquid having refractive index 1.75. Find the new focal length of the lens.
The focal length of the glass lens in air is given by1fa=(ang−1)(1R1−1R2)
Here, fa=20 cm and ang=1.5.
∴120cm=(1.5−1)(1R1−1R2)
(i) The focal length of the glass lens in liquid is given by
1fl=(wng−1)(1R1−1R2)
where lng is refractive index of glass with respect to the liquid.
Now, lng=anganl=1.51.75=67
∴1fl=(67−1)(1R1−1R2) (ii)
Dividing Eq. (i) by eq. (ii), we get
fl20 cm=0.5−(1/7)=−3.5
fl=−3.5 ×20 cm=−70 cm.
Or
Since, the focal length in the liquid comes out to be negative, the natural of the immersed lens is concave.