Question

For a spherical surface of radius of curvature R separating two media of refractive indices ${\mu }_1$ and ${\mu }_2$, the two principal focal lengths are $f_1$ and $f_2$ respectively. Which one of the following relations is correct?
A. $f_1$ = $f_2$
B. ${\displaystyle \frac{f_1}{{\mu }_2}}={\displaystyle \frac{f_2}{{\mu }_1}}$
C. ${\displaystyle \frac{f_1}{{\mu }_2}}=-{\displaystyle \frac{f_2}{{\mu }_1}}$
D. ${\displaystyle \frac{f_1}{{\mu }_1}}={\displaystyle \frac{f_2}{{\mu }_2}}$

Answer 1

Hint: This question can easily be solved using the formula for the relation for the refraction at a single spherical refracting surface. Use this relation to find the relation between the two sides of lenses with different refractive indices. Then, suppose the light is approaching the lens from the side with the refractive index ${\mu }_1$ and the focal length $f_1$. After the first refractions, the image will become the object for the second side and then solve again for the case.

Complete answer:
Before we start solving the question, let us take a look at all the parameters that have been given to us in the above question.
Radius of curvature = R
Two media of refractive indices ${\mu }_1$ and ${\mu }_2$, the two principal focal lengths are $f_1$ and $f_2$ respectively
Now,
For the surface having the focal length and refractive index $f_1$ and ${\mu }_1$respectively
So,
$\Rightarrow {\displaystyle \frac{\mu }{v_1}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{\mu -{\mu }_1}{R}}$ ……………. (1)
Now,
For the surface having the focal length and refractive index $f_2$ and ${\mu }_2$respectively
So,
$\Rightarrow {\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{\mu }{v_1}}={\displaystyle \frac{{\mu }_2-\mu }{R}}$ ……………. (2)
Now,
Adding the equations (1) and (2)
We have
$$\Rightarrow {\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{\mu -{\mu }_1}{R}}+{\displaystyle \frac{{\mu }_2-\mu }{R}}$$
$$\Rightarrow {\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{u}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$$

Now,
For the first refractive
$u=\mathrm{\infty }$
And, $v=f_1$
We have,
$$\Rightarrow {\displaystyle \frac{{\mu }_2}{f_1}}-{\displaystyle \frac{{\mu }_1}{\mathrm{\infty }}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$$
$$\Rightarrow {\displaystyle \frac{1}{f_1}}={\displaystyle \frac{1}{{\mu }_2}}[{\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}]$$ ………………….. (3)

For the second refractive
$v=\mathrm{\infty }$
And, $u=f_2$
We have,
$$\Rightarrow {\displaystyle \frac{{\mu }_2}{\mathrm{\infty }}}-{\displaystyle \frac{{\mu }_1}{f_2}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$$
$$\Rightarrow -{\displaystyle \frac{1}{f_2}}={\displaystyle \frac{1}{{\mu }_1}}[{\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}]$$ ………………….. (4)
Now, dividing equation (3) from equation (4)
We have
$$\Rightarrow {\displaystyle \frac{f_2}{f_1}}=-{\displaystyle \frac{{\mu }_1}{{\mu }_2}}$$
Or,
${\displaystyle \frac{f_1}{{\mu }_2}}=-{\displaystyle \frac{f_2}{{\mu }_1}}$

So, for a spherical surface of radius of curvature R separating two media of refractive indices ${\mu }_1$ and ${\mu }_2$, the two principal focal lengths are $f_1$ and $f_2$ respectively. Which one of the following relations is ${\displaystyle \frac{f_1}{{\mu }_2}}=-{\displaystyle \frac{f_2}{{\mu }_1}}$.

So, the correct answer is “Option C”.

Note:
Ibn al-Haitham is known as the father of optics and vision theory. During the 13th century, the first wearable glasses recognised by history emerged in Italy. Primitive glass-blown lenses were mounted into frames of wood or leather (or sometimes animal horn frames) and either carried in front of the face or perched on the nose.