Question

Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 Cm. What is the refractive index of the liquid?

Answer 1

Let focal length of convex lens (glass) , f1 = 30cm
focal length of plane concave lens (liquid) is f2
and it is given that combined focal length , F =45cm

we know,
1/f1 + 1/f2 = 1/F [ combined focal length formula]

so, 1/f2 = 1/F -1/f1

1/f2 = 1/45 - /1/30

= (2 - 3)/90 = -1/90

f2 = -90cm

for lens (glass), R1 = R and R2 = -R
use lens make formula,

1/f1 = (u - 1)[1/R1 -1/R2]

1/30 = (3/2 - 1) [ 1/R + 1/R ]

1/30 = 1/2 × 2/R = 1/R

R = 30cm

now, for plane concave lens (liquid)
R1 = -R = -30cm , R2 = ∞

so, 1/f2 = (u' - 1)[1/R1 - 1/R2]

1/-90 = (u' - 1)[1/-30 - 1/∞]

-1/90 = (u' - 1)/-30

1/3 = u' - 1

u' = 4/3

hence, refractive index of liquid is 4/3

Answer 2

Explanation:

It can be done in two ways.