a convex lens of refractive index 1.5 has a focal length of 20cm in air.calculate the change in focal length when it is immersed in water of refractive index 4/3.     

 

    pls answer this question

 

HEY LOOK.......USE LENS MAKER FORMULA............

1)  1/F =  (  u_l  -  1)  (  1/ R₁ -  1/R₂₎     ( F=20 ,  u_{l =} 1.5)

NOW WHEN PLACED IN LIQUID..................WE GET............

   2)1/f =   (u_{l^{/  u}w   ^{-  1)    ( 1 / R}1  ^{-   1/R}2)}

_{DIVIDE 1 BY 2..................}WE GET

F= 80 CM...................

I THINK SO.....IT IS RIGHT.............

THANKU

 

 

 

 

 

Dear

.  The solution is correcect