Question

Figure 9.37 shows an equiconvex lens (of refractive index 1.50) incontact with a liquid layer on top of a plane mirror. A small needlewith its tip on the principal axis is moved along the axis until itsinverted image is found at the position of the needle. The distance ofthe needle from the lens is measured to be 45.0cm. The liquid isremoved and the experiment is repeated. The new distance ismeasured to be 30.0cm. What is the refractive index of the liquid?

Answer 1

Given,

Focal length of the convex lens f 1 =30 cm.

Combined focal length of the system is 45 cm

The combined focal length of the system is,

1 f = 1 f 1 + 1 f 2

Substitute the given values.

1 45 = 1 30 + 1 f 2

1 f 2 = 1 45 − 1 30

1 f 2 = −15 45×30

f 2 =−90 cm

Let the radius of curvature of the surface is R.

Thus, the radius of curvature of the other surface will be −R

Thus,

1 f 1 =( μ−1 )( 1 R − 1 − R )

1 30 =( 1.5−1 )( 2 R )

R=0.5×2×30

R=30 cm

Let μ 2 be the reflective index of the liquid.

Since, radius of curvature of the liquid on the side of the plane mirror will be infinite and radius of curvature of the liquid on the side of the lens is −R. Thus, the focal length of the liquid is,

1 f 2 =( μ 2 −1 )( 1 −R − 1 ∞ )

1 −90 =( μ 2 −1 )( 1 −30 −0 )

( μ 2 −1 )= 1 3

μ 2 =1.33

Thus, the reflective index of the liquid is 1.33.