Question

A double concave lens with refractive index 1.6is kept in air whose R1 and R2 are 60 and 120 then it's focallength

Answer 1

1/f =( n2/n1 - 1 )×(1/R1 -1/R2)

n1 =1

n2 = 1.6

R1 = -60

R2 = 120

therefore

1/f = (1.6-1)×(1/-60 -1/120)

= 0.6×(-180/60×120)

1/f = - 0.3/20

f = - 20/0.3

f = - 66.66