Question

Two similar equiconvex lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_1$ . When the space between the two lenses is filled with glycerine, (which has the same refractive index $(\mu =1.5)$ as that of glass) then the equivalent focal length is $F_2$ . the ratio $F_1:F_2$ will be:
A) $2:1$
B) $1:2$
C) $2:3$
D) $3:4$

Answer 1

Hint: First, calculate the focal length of both the given lenses individually, and then use it to calculate the value of $F_1$ . Then consider the glycerine as a concave lens between the convex lenses and calculate its focal length. And then use all three focal lengths to calculate the value of $F_2$ . Divide the two to get the ratio.

Formula Used:
Focal length of a lens is given by, ${\displaystyle \frac{1}{f}}=({\displaystyle \frac{{\mu }_L}{{\mu }_S}}-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$ where, ${\mu }_L$ is the refractive index of lens, ${\mu }_S$ is the refractive index of surroundings of the lens, $R_1$ is the radius of curvature of one side of the lens, $R_2$ is the radius of curvature of second side of the lens and $f$ is the focal length of the lens.

Complete Step by Step Solution:
First of all, we will calculate the focal lengths of each of the equiconvex lenses given in the question.
For lens 1, let the focal length be $f_1$
Therefore, ${\displaystyle \frac{1}{f_1}}=({\displaystyle \frac{{\mu }_L}{{\mu }_S}}-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$
Refractive index of glass is $1.5$ , as given in the question, the refractive index of air is 1, since we are given an equiconvex lens, its radius of curvature will be equal for both sides. Let it be $R$
Therefore, we get ${\displaystyle \frac{1}{f_1}}=({\displaystyle \frac{1.5}{1}}-1)({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{-R}})$
(since the two sides of the lens will have radius of curvatures in opposite direction, we put the negative sign to indicate it)
On solving, we get ${\displaystyle \frac{1}{f_1}}=0.5\times {\displaystyle \frac{2}{R}}={\displaystyle \frac{1}{R}}$
$\Rightarrow f_1=R$
The other lens is similar, as given in question. We will apply the same method to calculate its focal length. Let its focal length be $f_2$ . therefore, $f_2=R$
Now, when these two are put coaxially, the resultant focal length is given to be $F_1$ . therefore,
${\displaystyle \frac{1}{F_1}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}$
On putting values, ${\displaystyle \frac{1}{F_1}}={\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}}={\displaystyle \frac{2}{R}}$
$\Rightarrow F_1={\displaystyle \frac{R}{2}}$
Now, when glycerine is put between these equiconvex lenses, it will gain the shape of a concave lens as glycerine is a liquid and liquids take the shape of its surroundings.
So, imagine it as a concave lens made of glycerine and calculate its focal length.
Let its focal length be $f_3$ therefore, ${\displaystyle \frac{1}{f_3}}=({\displaystyle \frac{{\mu }_L}{{\mu }_S}}-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$
The concave lens will have similar radii of curvature for its both sides. However, its first side will have a negative radius of curvature and the second side will have a positive radius of curvature, with respect to the previous considerations. Let the radius of curvature of this lens have a value $R$
Therefore, ${\displaystyle \frac{1}{f_3}}=({\displaystyle \frac{1.5}{1}}-1)({\displaystyle \frac{1}{-R}}-{\displaystyle \frac{1}{R}})$ (refractive index of glycerine is given)
$\Rightarrow {\displaystyle \frac{1}{f_3}}=0.5\times {\displaystyle \frac{-2}{R}}={\displaystyle \frac{-1}{R}}$
$\Rightarrow f_3=-R$
Now, when these three lenses are in combination, we get the resultant focal length as $F_2$
Therefore, $${\displaystyle \frac{1}{F_2}}={\displaystyle \frac{1}{f_1}}+{\displaystyle \frac{1}{f_2}}+{\displaystyle \frac{1}{f_3}}$$
$${\displaystyle \frac{1}{F_2}}={\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{R}}={\displaystyle \frac{1}{R}}$$
$\Rightarrow F_2=R$
Hence, $F_1:F_2={\displaystyle \frac{R}{2}}:R=1:2$

Option (B) is the correct answer.

Note: While calculating focal length, make sure you put the appropriate sign before radii of curvature. Also, consider the same sign notations in the entire question. For example, if you take the left side as negative, then use the left side as negative only for the rest of your question. Mixing up the sign notation will result in an incorrect answer.