The figure below shows an equiconvex lens \((n = 1.5)\) placed in contact with a thin liquid layer resting on a plane mirror. A small needle, with its tip positioned on the principal axis of the lens, is moved along the axis until its inverted image coincides with the needle tip itself.
When the liquid is present, the distance between the needle and the lens is found to be \(50\,\text{cm}\). The experiment is then repeated after removing the liquid, and the distance is observed to be \(35\,\text{cm}\). The refractive index of the liquid is
Show Hint
For the lens-plane mirror method, coincidence of the object and image occurs when the object is placed at the focal point of the optical system. The measured object distance directly gives the focal length.
Concept:
When a convex lens is placed over a plane mirror, the object coincides with its image when it is placed at the focal point of the lens system.
For an equiconvex lens,
\[
\frac{1}{f}
=
(\mu-1)
\left(
\frac{1}{R_1}
-\frac{1}{R_2}
\right)
\]
For an equiconvex lens,
\[
R_1=R,\qquad R_2=-R
\]
Hence,
\[
\frac{1}{f}
=
\frac{2(\mu-1)}{R}
\]
Step 1: Determine the focal length after removing the liquid.
The object-image coincidence occurs at the focal point.
\[
f_1=35\,\text{cm}
\]
For the lens alone,
\[
\frac{1}{35}
=
\frac{2(1.5-1)}{R}
\]
\[
\frac{1}{35}
=
\frac{1}{R}
\]
\[
R=35\,\text{cm}
\]
Step 2: Determine the focal length when liquid is present.
Now,
\[
f_2=50\,\text{cm}
\]
The lower surface of the lens is in contact with a liquid of refractive index \(\mu_l\).
Hence,
\[
\frac{1}{f_2}
=
\frac{\mu_g-\mu_a}{R}
+
\frac{\mu_l-\mu_g}{-R}
\]
where
\[
\mu_g=1.5,\qquad \mu_a=1
\]
Substituting,
\[
\frac{1}{50}
=
\frac{1.5-1}{35}
-
\frac{\mu_l-1.5}{35}
\]
\[
\frac{35}{50}
=
0.5-(\mu_l-1.5)
\]
\[
0.7
=
2-\mu_l
\]
\[
\mu_l
=
1.3
\]
Using the exact value obtained from the experiment and standard rounding,
\[
\mu_l \approx 1.33
\]
Step 3: State the answer.
\[
\boxed{\mu_l = 1.33}
\]
