Question

A thin lens made of a material of refractive index μ₂ has a medium of refractive index μ₁ on one side and a medium of refractive index μ₃ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium μ₁ and (b) from the medium μ₃?

Answer 1

Given,
A biconvex lens with two radii of curvature that have equal magnitude R.
Refractive index of the material of the lens is μ₂.
First medium of refractive index is μ_1.
Second medium of refractive index is μ₃.
As per the question, the light beams are travelling parallel to the principal axis of the lens.
i.e., u (object distance) = ∞

(a) The light beam is incident on the lens from first medium μ₁.
Thus, refraction takes place at first surface
Using equation of refraction,

$\frac{{\mathrm{\mu }}_2}{v}-\frac{{\mathrm{\mu }}_1}{u}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{R}$
Where, v is the image distance.
Applying sign convention, we get:
$$\begin{gathered} \frac{{\mathrm{\mu }}_2}{v}-\frac{{\mathrm{\mu }}_1}{\left(-\infty \right)}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{R} \\ \Rightarrow \frac{1}{v}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{{\mathrm{\mu }}_{2}R} \\ \Rightarrow \Rightarrow v=\frac{{\mathrm{\mu }}_{2}R}{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1} \end{gathered}$$

Now, refraction takes place at 2^nd surface

Thus,$\frac{{\mathrm{\mu }}_3}{v}-\frac{{\mathrm{\mu }}_2}{u}=\frac{{\mathrm{\mu }}_3-{\mathrm{\mu }}_2}{R}$
Here, the image distance of the previous case becomes object distance:

$$\begin{gathered} \frac{{\mathrm{\mu }}_3}{v}=-\left[\frac{{\mathrm{\mu }}_3-{\mathrm{\mu }}_2}{R}-\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_{2}R}\left({\mathrm{\mu }}_2-{\mathrm{\mu }}_1\right)\right] \\ =-\left[\frac{{\mathrm{\mu }}_3-{\mathrm{\mu }}_2-{\mathrm{\mu }}_2+{\mathrm{\mu }}_1}{R}\right] \\ v=-\left[\frac{{\mathrm{\mu }}_{3}R}{{\mathrm{\mu }}_3-2{\mathrm{\mu }}_2+{\mathrm{\mu }}_1}\right] \end{gathered}$$
Therefore, the image is formed at $\frac{{\mathrm{\mu }}_{3}R}{2{\mathrm{\mu }}_2-{\mathrm{\mu }}_1-{\mathrm{\mu }}_3}$

(b) The light beam is incident on the lens from second medium μ₃.
Thus, refraction takes place at second surface.
Using equation of refraction,
$\frac{{\mathrm{\mu }}_2}{v}-\frac{{\mathrm{\mu }}_3}{u}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_3}{R}$
Where, v is the image distance
Applying sign convention, we get:

$$\begin{gathered} \frac{{\mathrm{\mu }}_2}{v}-\frac{{\mathrm{\mu }}_3}{\left(-\infty \right)}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_3}{R} \\ \Rightarrow \frac{1}{v}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_3}{{\mathrm{\mu }}_{2}R} \\ \Rightarrow \Rightarrow v=\frac{{\mathrm{\mu }}_{2}R}{{\mathrm{\mu }}_2-{\mathrm{\mu }}_3} \end{gathered}$$
Now, refraction takes place at 2nd surface.

Thus,$\frac{{\mathrm{\mu }}_1}{v}-\frac{{\mathrm{\mu }}_2}{u}=\frac{{\mathrm{\mu }}_1-{\mathrm{\mu }}_2}{R}$

Here, the image distance of the previous case becomes object distance.
$$\begin{gathered} \frac{{\mathrm{\mu }}_1}{v}=-\left[\frac{{\mathrm{\mu }}_1-{\mathrm{\mu }}_2}{R}-\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_{2}R}\left({\mathrm{\mu }}_2-{\mathrm{\mu }}_3\right)\right] \\ =-\left[\frac{{\mathrm{\mu }}_1-{\mathrm{\mu }}_2-{\mathrm{\mu }}_2+{\mathrm{\mu }}_3}{R}\right] \\ v=-\left[\frac{{\mathrm{\mu }}_{1}R}{{\mathrm{\mu }}_3-2{\mathrm{\mu }}_2+{\mathrm{\mu }}_1}\right] \end{gathered}$$
Therefore, the image is formed at $\frac{{\mathrm{\mu }}_{1}R}{2{\mathrm{\mu }}_2-{\mathrm{\mu }}_1-{\mathrm{\mu }}_3}$