OPTICS OBJECTIVE 9

Solution 9

Problem 9: Achromatic Doublet Radius

Step-by-Step Derivation

1. Condition for Achromatism:
For a combination of two lenses to be achromatic (focal length independent of wavelength), the derivative of total power with respect to wavelength must be zero: $$ \frac{d}{d\lambda}\left( \frac{1}{F} \right) = \frac{d}{d\lambda}\left( \frac{1}{f_A} + \frac{1}{f_B} \right) = 0 $$

Using the Lens Maker’s Formula $\frac{1}{f} = (\mu – 1)K$, where $K$ is the geometric factor $(\frac{1}{R_1} – \frac{1}{R_2})$, we get: $$ \frac{d}{d\lambda} [(\mu_A – 1)K_A + (\mu_B – 1)K_B] = 0 $$ $$ K_A \frac{d\mu_A}{d\lambda} + K_B \frac{d\mu_B}{d\lambda} = 0 $$ Given $\mu = \mu_0 – \beta(\lambda – \lambda_0)$, the derivative is $\frac{d\mu}{d\lambda} = -\beta$. Thus: $$ K_A (-\beta_A) + K_B (-\beta_B) = 0 \implies K_A \beta_A + K_B \beta_B = 0 $$

2. Geometric Factors ($K$):
Let $R$ be the radius of the common surface.

Lens A (outer $r_1$, common $R$): $K_A = \left( \frac{1}{r_1} – \frac{1}{R} \right)$
Lens B (common $R$, outer $r_2$): $K_B = \left( \frac{1}{R} – \frac{1}{r_2} \right)$

3. Solving for $R$:
Substitute $K_A$ and $K_B$ into the condition: $$ \beta_A \left( \frac{1}{r_1} – \frac{1}{R} \right) + \beta_B \left( \frac{1}{R} – \frac{1}{r_2} \right) = 0 $$ $$ \frac{\beta_A}{r_1} – \frac{\beta_A}{R} + \frac{\beta_B}{R} – \frac{\beta_B}{r_2} = 0 $$ $$ \frac{1}{R} (\beta_B – \beta_A) = \frac{\beta_B}{r_2} – \frac{\beta_A}{r_1} $$ $$ \frac{1}{R} (\beta_B – \beta_A) = \frac{\beta_B r_1 – \beta_A r_2}{r_1 r_2} $$ $$ R = \frac{r_1 r_2 (\beta_B – \beta_A)}{\beta_B r_1 – \beta_A r_2} $$ Multiply numerator and denominator by -1 to match the options: $$ R = \frac{r_1 r_2 (\beta_A – \beta_B)}{\beta_A r_2 – \beta_B r_1} $$