Question

Focal length of thin lens in air is 10cm . Medium on one side of lens is replaced by a medium of refractive index 2 .The radius of curvature of surface of lens in contact with the medium is 20cm find new focal length​

Answer 1

since the radius of curvature is 20cm of the half lens and the radius of curvature of the other half is 5cm

so

New radius of curvature is 25cm

then

focal length is 50cm

Answer 2

Answer:40cm

Explanation: let n be the refractive index of lens

Applying Retraction Formula:

  at 1st surface:- n/v1 - 0 = n-1/R1

  at 2nd surface:- 2/v - n/v1 = 2-n/(-20)

Adding these two eqns:

2/v= n-1/R1 + 2-n/(-20)

    = n-1/R1 + 1-n/(-20) + 1/(-20)

    = (n-1)[ 1/R1 - 1/(-20)] +1/(-20)

    = 1/f + 1/(-20)...... (Here f is initial focal length)

    = 1/10 - 1/20

    = 1/20

2/v= 1/20

2/f= 1/20....... (Since final image will be at f as incident Rays on 1st surface are considered parallel)

2×20=1×f

Therefore,

f=40 cm.