Question

A thin convex lens with its two surfaces of radii of curvature $R_1$ and $R_2$ is made of a material of refractive index $\mu_2$. A medium of refractive index $\mu_1$ ($< \mu_2$) surrounds it. Using a ray diagram, derive the Lens Maker's formula when a point object placed on the principal axis in front of the surface of radius of curvature $R_1$ produces an image I on the other side of the lens.

Hint:

Apply the single-surface refraction formula twice, once at each surface, add the two equations and put the object at infinity.

Answer 1

Step 1: Ray diagram in words:
Draw a thin convex lens of refractive index \(\mu_2\) with surface 1 of radius \(R_1\) (centre \(C_1\)) and surface 2 of radius \(R_2\) (centre \(C_2\)). The surrounding medium has refractive index \(\mu_1\).
Place a point object O on the principal axis. A ray from O strikes surface 1 and bends towards the normal. If the second surface were absent, it would form an image \(I_1\) inside the lens. The second surface then bends the ray again to meet the axis at the final image I.

Step 2: Refraction at a single spherical surface:
Let a ray from O meet a spherical surface at A, with centre of curvature C and pole P. Let M be the foot of the perpendicular from A on the axis. The refracted ray meets the axis at I.
Angles: \(\alpha = \angle AOP\), \(\beta = \angle AIP\), \(\gamma = \angle ACP\). From the exterior angles of triangles AOC and ACI: \(i = \alpha + \gamma\) and \(r = \gamma - \beta\).
For small angles Snell's law gives \(\mu_1 i = \mu_2 r\), so \(\mu_1(\alpha + \gamma) = \mu_2(\gamma - \beta)\).
For a ray close to the axis: \(\alpha \approx AM/MO\), \(\beta \approx AM/MI\), \(\gamma \approx AM/MC\), and M is almost at P. Substituting and using signs (\(PO = -u\), \(PI = +v\), \(PC = +R\)):
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]

Step 3: Apply to the first surface:
The object O is at distance u from the lens (pole P). Surface 1 forms an image \(I_1\) at distance \(v_1\) in the medium \(\mu_2\):
\[ \frac{\mu_2}{v_1} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1} \quad \cdots (1) \]

Step 4: Apply to the second surface:
For surface 2, \(I_1\) acts as the object, and light goes from \(\mu_2\) to \(\mu_1\). The lens is thin, so distances from the two poles are almost equal. The final image is at v:
\[ \frac{\mu_1}{v} - \frac{\mu_2}{v_1} = \frac{\mu_1 - \mu_2}{R_2} \quad \cdots (2) \]

Step 5: Add and simplify:
Adding (1) and (2) cancels \(\mu_2/v_1\):
\[ \frac{\mu_1}{v} - \frac{\mu_1}{u} = (\mu_2 - \mu_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Divide by \(\mu_1\):
\[ \frac{1}{v} - \frac{1}{u} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
For \(u = -\infty\), \(v = f\). This gives the Lens Maker's formula.

Final Answer:
The focal length of the thin lens in a medium is given by: \[ \boxed{\frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)} \]