A convex lens have focal length of 20 cm. the material of lens has refractive index of 1.5 is immersed in water of refractive index 4/3. find the change in focal length of the lens.
1fa=(ang−1)(1R1−1R2)−−−(1)
give that ang=1.5=refrentiue indent of gases w.r.t air
fa= focal length in air
⇒ ft= focal length in liquid will be given
1ft=(lng−1)(1R1−1R2)−−−(2)
dividing (1) & (2) we get,
ftfa=(ang−1)(lng−1)=(1.5−1)(anganl)=0.5(3/24/3−1)
⇒ fl=fa×0.5×8=20×4=80 cm
charge =fl−fa=80−20=60 cm