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 $45.0\text{cm}$. The liquid is removed, and the experiment is repeated. The new distance is measured to be $30.0\text{cm}$. What is the refractive index of the liquid?

Answer 1

Step 1: Find the focal length of the concave lens of liquid.
Formula Used: $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
In the absence of liquid, for the combination of convex lens and plane mirror, inverted image will coincide with the object only when the object is placed at focus of convex lens. Same rule applies in the presence of the liquid between lens and the mirror. Hence the object or image distance given can be taken as focal length of the respective combinations.

When the liquid is used, the system acts as a combination of convex lens and a concave lens.
Therefore, the focal length of the combination is f (say) $=45\text{cm}$
When the liquid is removed, the image is formed at $30\text{cm}$.
Therefore, the focal length of convex lens, $f_1=30\text{cm}$
Let the focal length of the concave lens of liquid be $f_2$.

For a pair of optical systems placed in contact, the equivalent focal length is given by,
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
Putting the values, we get,
$\frac{1}{45}=\frac{1}{30}+\frac{1}{f_2}$
$\frac{1}{f_2}=\frac{1}{45}-\frac{1}{30}$
$\frac{1}{f_2}=\frac{2-3}{90}=-\frac{1}{90}$
$f_2=-90\text{cm}$

Step 2: Find the value of radius of curvature.
Hint: Lens maker’s formula: $\frac{1}{f}=(n_{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
Now, let the refractive index of the lens be $n_1$
and the radius of curvature of one surface be $R_1=R.$
Then, the radius of curvature of the other surface is $R_2=-R$

Using the lens maker,s formula,
$\frac{1}{f_1}=(n_1-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{30}=(1.5-1)(\frac{1}{R}-\frac{1}{-R})$
$\frac{1}{30}=0.5\left(\frac{2}{R}\right)$
$\frac{1}{30}=\frac{1}{R}$
$R=30\text{cm}$.

Step 3: Find the refractive index of the liquid.
Hint: Lens maker’s formula.
Formula Used: $\frac{1}{f}=(n_{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
Radius of curvature of the liquid lens, $R_1^{\prime }=-30\text{cm}$.
The other side of the liquid is plane, therefore, $R_2^{\prime }=\infty$
Let $n_2$ be the refractive index of the liquid.

Using the lens maker's formula,
$\frac{1}{f_2}=(n_2-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$-\frac{1}{90}=(n_2-1)(\frac{1}{-30}-\frac{1}{\infty })$
$-\frac{1}{90}=-\frac{1}{30}(n_2-1)$
$n_2-1=\frac{1}{3}$
$n_2=1.33$
Hence, the refractive index of the liquid is 1.33.
Final Answer: $n=1.33$