Question

A concave lens of f=50 is made of material of nr for red lights is 1.640 and nb for
blue light is 1.658.It is then combined with a convex lens of dispersive power
0.0172 to form an achromatic doublet, determination the focal length of the
achromatic lens

Answer 1

Answer on Question #69251, Physics / Optics

Question. A concave lens of $f = 50 \, \text{cm}$ is made of material of $n_r$ for red lights is 1,640 and $n_b$ for blue light is 1,658. It is then combined with a convex lens of dispersive power 0,0172 to form an achromatic doublet, determination the focal length of the achromatic lens.

Given.

focal length of concave lens: $f_1 = -50 \, \text{cm} = -0.5 \, \text{m}$;

refractive index for red lights: $n_r = 1,640$;

refractive index for blue lights: $n_b = 1,658$;

dispersive power of a convex lens: $\omega_2 = 0.0172$.

Find.

focal length of the achromatic lens: $f$.

Solution.

In accordance with the conditions for achromatic doublet of two lens

$$P = P_1 + P_2;$$ $$\omega_1 P_1 + \omega_2 P_2 = 0,$$

where $P$ – optical power of the achromatic doublet; $P_1 = \frac{1}{f_1}, P_2 = \frac{1}{f_2}$ – optical power of the first and second lens; $\omega_1, \omega_2$ – dispersive power of the first and second lens.

The expression for dispersive power of the material of the thin lens

$$\omega_1 = \frac{n_b - n_r}{n - 1}.$$

Generally, $n$ is replaced by the mean value of $n_b$ and $n_r$ which is

$$n = \frac{n_b + n_r}{2}.$$

Then

$$\omega_1 / f_1 + \omega_2 / f_2 = 0,$$ $$\frac {\omega_ {1}}{f _ {1}} = - \frac {\omega_ {2}}{f _ {2}}, f _ {2} = - \frac {\omega_ {2}}{\omega_ {1}} f _ {1} = - \frac {\omega_ {2}}{\frac {n _ {b} - n _ {r}}{n - 1}} f _ {1} = - \frac {\omega_ {2} (n - 1)}{n _ {b} - n _ {r}} f _ {1} = - \frac {\omega_ {2} \left(\frac {n _ {b} + n _ {r}}{2} - 1\right)}{n _ {b} - n _ {r}} f _ {1} = - \frac {\omega_ {2} \omega_ {1} (n - 1)}{n - 1} f _ {1} = - \frac {\omega_ {2} \omega_ {2} (n - 1)}{n - 2} f _ {1} = - \frac {\omega_ {2} (n _ {b} + n _ {r} - 2)}{2 (n _ {b} - n _ {r})} f _ {1} = - \frac {0 , 0 1 7 2 (1 , 6 5 8 + 1 , 6 4 0 - 2)}{2 (1 , 6 5 8 - 1 , 6 4 0)} \cdot (- 0, 5) = 0, 6 2 m.$$

The focal length of the achromatic doublet

$$\frac {1}{f} = \frac {1}{f _ {1}} + \frac {1}{f _ {2}}; f = \frac {f _ {1} f _ {2}}{f _ {1} + f _ {2}} = \frac {- 0 , 5 \cdot 0 , 6 2}{- 0 , 5 + 0 , 6 2} = - 2, 5 8 m$$

Answer: The focal length of the achromatic lens: $f = -2,58 \, m$ .

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