Question

Whrn an transparent convex lens of refractive index 1.33 is immersed in water then what is effect on its focal lenghth

Answer 1

Refractive index of glass w.r.t. air = μg
Refractive index of the liquid w.r.t. air = μl = 2μg
If a lens made of glass of refracive index μg is immersed in a liquid of refractive index μl then its focal length fl is given by
1/fl = (l μg − 1) (1/R1 − 1/R2)
If f is the focal length of the lens in air then
1/f = (aμg − 1) (1/R1 − 1/R2)
Therefore
fl/f = (aμg − 1)/(l μg − 1)
= (a μg − 1) / (a μg / a μl − 1)
= (μg − 1) / (μg /μl − 1)

The liquid is water.
μl = μw = 1.33
μg ≈ 1.5 > μw
fw/f = (1.5 − 1) / (1.5 /1.33− 1)

= 0.5/(1.13 − 1)
= 3.846
fw = 3.846 f

The focal length increases in water.