Find the change in focal length of a convex lens of focal length 20cm when it is immersed in water. The refractive indices of glass and water are 3/2 and 4/3 respectively.
Dear Student,
1/f = (ng-nm)(1/R1-1/R2)
f= foal length of the lens
ng= refractive index of the glass=1.5
nm= refrctive index of the medium
R1&R2 are the radius of curvature near and far from the sorce respectively.
in air , refractive index of the air is 1.
Hence, the focal length in air is given by :-
1/fa = (1.5-1)(1/R1-1/R2) ------(1)
in water, refractive index of the water is 1.33.
Hence, the focal length in water is given by :
1/fw = (1.5-1.33)(1/R1-1/R2) -----(2)
divide equation no. (1) and (2), we get
fw/fa = (1.5-1)/(1.5-1.33)
= .5/.17
= 2.94
Given :- fa = 20cm
fw = 20*2.94
=58.80cm
Hence, focal length of the lens in the water in 58.80cm.
1/f = (ng-nm)(1/R1-1/R2)
f= foal length of the lens
ng= refractive index of the glass=1.5
nm= refrctive index of the medium
R1&R2 are the radius of curvature near and far from the sorce respectively.
in air , refractive index of the air is 1.
Hence, the focal length in air is given by :-
1/fa = (1.5-1)(1/R1-1/R2) ------(1)
in water, refractive index of the water is 1.33.
Hence, the focal length in water is given by :
1/fw = (1.5-1.33)(1/R1-1/R2) -----(2)
divide equation no. (1) and (2), we get
fw/fa = (1.5-1)/(1.5-1.33)
= .5/.17
= 2.94
Given :- fa = 20cm
fw = 20*2.94
=58.80cm
Hence, focal length of the lens in the water in 58.80cm.