Question

A thin lens of focal length + 12 cm is immersed in water `(mu = 1.33).` What is its new focal length ?

Answer 1

`f_a=12cm`,`_au_g=(3)/(2)`,`._au_(omega)=(4)/(3)`
`(1)/(f_a)=(._au_g-1)((1)/(R_1)-(1)/(R_2))` …(i)
`(1)/(f_(omega))=(._(omega)u_g-1)((1)/(R_1)-(1)/(R_2))` ..(ii)
`(i)//(ii)`
`(f_(omega))/(f_a)=((._au_(g)-1))/((._(omega)u_(g)-1))=((._au_(g)-1))/(((._au_(g))/(._au_(omega))-1))=((3//2)-1)/(((3//2)/(4//3)-1))=(((1)/(2)))/(((1)/(8)))=4`
`f_(omega)=4f_(a)=4xx12=48cm`