A thin lens made of a material of refractive index $\mu_2$ has a medium of refractive index $\mu_1$ on one side and a medium of refractive index $\mu_3$ 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:

The medium $\mu_1$

From the medium $\mu_3?$

Q: A thin lens made of a material of refractive index $\mu_2$ has a medium of refractive index $\mu_1$ on one side and a medium of refractive index $\mu_3$ 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: 1. The medium $\mu_1$ 2. From the medium $\mu_3?$

1. When the beam is incident on the lens from medium $\mu_{3}$ Then $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$ or $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{(-\infty)}=\frac{\mu_2-\mu_1}{\text{R}}$ or $\frac{1}{\text{v}}=\frac{\mu_2-\mu_1}{\mu_2\text{R}}$ or $\text{v}=\frac{\mu_2\text{R}}{\mu_2-\mu_1}$ Again, for $2^{nd} $ refraction, $\frac{\mu_3}{\text{v}}-\frac{\mu_2}{\text{u}}=\frac{\mu_3-\mu_2}{\text{R}}$ or, $\frac{\mu_3}{\text{v}}=-\Big[\frac{\mu_3-\mu_2}{\text{R}}-\frac{\mu_2}{\mu_2\text{R}}(\mu_2-\mu_1)\Big]\Rightarrow-\Big[\frac{\mu_3-\mu_2-\mu_2+\mu_1}{\text{R}}\Big]$ or, $\text{v}=-\Big[\frac{\mu_3\text{R}}{\mu_3-2\mu_2+\mu_1}\Big]$ So, the image will be formed at $=\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$ 2. Similarly for the beam from $\mu_3$ medium the image is formed at $\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

Question

A thin lens made of a material of refractive index $\mu_2$ has a medium of refractive index $\mu_1$ on one side and a medium of refractive index $\mu_3$ 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:

The medium $\mu_1$
From the medium $\mu_3?$

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Solution

When the beam is incident on the lens from medium $\mu_{3}$

Then $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$ or $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{(-\infty)}=\frac{\mu_2-\mu_1}{\text{R}}$
or $\frac{1}{\text{v}}=\frac{\mu_2-\mu_1}{\mu_2\text{R}}$ or $\text{v}=\frac{\mu_2\text{R}}{\mu_2-\mu_1}$
Again, for $2^{nd} $ refraction, $\frac{\mu_3}{\text{v}}-\frac{\mu_2}{\text{u}}=\frac{\mu_3-\mu_2}{\text{R}}$
or, $\frac{\mu_3}{\text{v}}=-\Big[\frac{\mu_3-\mu_2}{\text{R}}-\frac{\mu_2}{\mu_2\text{R}}(\mu_2-\mu_1)\Big]\Rightarrow-\Big[\frac{\mu_3-\mu_2-\mu_2+\mu_1}{\text{R}}\Big]$
or, $\text{v}=-\Big[\frac{\mu_3\text{R}}{\mu_3-2\mu_2+\mu_1}\Big]$
So, the image will be formed at $=\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

Similarly for the beam from $\mu_3$ medium the image is formed at $\frac{\mu_3\text{R}}{2\mu_2-\mu_1-\mu_3}$

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