Focal length of an equiconvex lens is 1.00 m when it is placed under water calculate the focal length of same lens in air take refractive indices of water and glass as 4/3 and 3/2 respectively
Hi Ashmita dear, recall the lens maker's formula
1/f = (u - 1) ( 1/R1 + 1/R2)
If we use equiconvex lens then R1 = R2 = R
So 1/f = (u-1) * 2/R
Here u is for mu which stands for refractive index of the material of the lens with respect to air medium. And it is given u = 3/2
Plugging we get f = R and so Radius of curvature is 1 m
Now as we place the lens immersed within water the refractive index of glass with respect to the surrounding medium will be (3/2)/(4/3)
So relative refractive index u1 = 9/8
Plugging 1/f1 = (9/8 -1) * 2/1 = 2/8 = 1/4
Hence new focal length would become 4 m
1/f = (u - 1) ( 1/R1 + 1/R2)
If we use equiconvex lens then R1 = R2 = R
So 1/f = (u-1) * 2/R
Here u is for mu which stands for refractive index of the material of the lens with respect to air medium. And it is given u = 3/2
Plugging we get f = R and so Radius of curvature is 1 m
Now as we place the lens immersed within water the refractive index of glass with respect to the surrounding medium will be (3/2)/(4/3)
So relative refractive index u1 = 9/8
Plugging 1/f1 = (9/8 -1) * 2/1 = 2/8 = 1/4
Hence new focal length would become 4 m
