Question

An equiconvex lens is placed in contact with plane mirror as shown in the. A small needle with its tip on principal axis is moved along the axis until its inverted image is found at the position of needle. The distance of the needle from the lens is measured to be 30cm. Now a few drops of liquid are put in between the lens and the plane mirror and the new position of the image is found to be located at 45cm. Calculate the refractive index of the liquid.

Answer 1

Given :

Focal length of lens f1=30cm

Focal length of liquid as it acts like mirror =f2

Foal length of system (lens+liquid]=45cm

Equivalength focal length is :

1/f=1/f1+1/f2

1/f2=1/f-1/f1

=1/45-1/30=2-3/90=-1/90

Therefore f2=-90cm
Let n1 be the refractive index of lens

R be the radius of curvature (first surface)

Hence radius of curvature of other surface= -R

from lens makers formula:1/f1=(n1-1)(1/R-1/-R)

1/30=(1.5-1)(2/R)

R=30/0.5x2=30cm

let n2 be the refractive index of liquid 

Radius of curvature of planemirror=inifinity=∞

Radius of curvature of liquid on side of lens R=-30cm

By lens makers formula:

1/f2=[n1-1](-1/R-1/∞]

-1/90=(n2-1)[1/30-0]

n2-1=30/90

n2-1=1/3

n2=1+1/3=4/3=1.33