CBSE 2025, Set 3

Question

A convex lens of focal length $10	ext{ cm}$, a concave lens of focal length $15	ext{ cm}$ and a third lens of unknown focal length are placed coaxially in contact. If the focal length of the combination is $+12	ext{ cm}$, find the nature and focal length of the third lens, if all lenses are thin. Will the answer change if the lenses were thick?

Edvaya · Accepted Answer

The correct answer is Subjective. Given:
* Focal length of convex lens, $f_1 = +10	ext{ cm}$
* Focal length of concave lens, $f_2 = -15	ext{ cm}$
* Focal length of combination, $F = +12	ext{ cm}$
* Let the focal length of the third lens be $f_3$.

For thin lenses in contact, the equivalent focal length is given by:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$$

Substituting the given values:
$$\frac{1}{12} = \frac{1}{10} + \frac{1}{-15} + \frac{1}{f_3}$$
$$\frac{1}{12} = \frac{1}{10} - \frac{1}{15} + \frac{1}{f_3}$$

Rearranging for $\frac{1}{f_3}$:
$$\frac{1}{f_3} = \frac{1}{12} - \frac{1}{10} + \frac{1}{15}$$
Taking the LCM of 12, 10, and 15, which is 60:
$$\frac{1}{f_3} = \frac{5 - 6 + 4}{60}$$
$$\frac{1}{f_3} = \frac{3}{60} = \frac{1}{20}$$

$$\boxed{f_3 = +20	ext{ cm}}$$

**Nature:** Since the focal length is positive, the third lens is a **Convex lens**.

**Thick lenses:** Yes, the answer will change if the lenses were thick. The simple formula $\frac{1}{F} = \sum \frac{1}{f_i}$ assumes the optical centers coincide, which is only valid for geometrically thin lenses. For thick lenses, the separation between their principal planes must be accounted for.

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Physics > Optics > Ray Optics And Optical Instruments

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