# A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal length is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths (in cm). A: plus 50, minus 75 B: plus 10, minus 15 C: minus 50, plus 75 D: minus 10, minus 15

## Question

A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal length is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths (in cm).

A: +50, -75 B: +10, -15 C: -50, +75 D: -10, -15

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## Answer

The correct answer is **B**.

To determine the individual focal lengths of the lenses, let's consider the well-known lens formula for lenses in contact:

$$ \frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} $$

where $f_{\text{eq}}$ is the equivalent focal length of the combined lens system.

Given:

- The
**equivalent focal length**of the combined lenses, $f_{\text{eq}} = 30$ cm. - The
**magnitude of the ratio**of their focal lengths is $2/3$.

Assume:

- The focal length of the
**convex lens**($ f_1 $) is $2f$, making the convex lens more powerful. - The focal length of the
**concave lens**($ f_2 $) is $-3f$, considering the focal length of concave lens to be negative as it diverges light.

Using the formula, we can write:

$$ \frac{1}{30} = \frac{1}{2f} + \frac{1}{-3f} $$

To solve for $f$, find a common denominator:

$$ \frac{1}{2f} + \frac{1}{-3f} = \frac{3 - 2}{6f} = \frac{1}{6f} $$

This simplifies to:

$$ \frac{1}{30} = \frac{1}{6f} $$

Solving for $f$, we get:

$$ f = 5 \text{ cm} $$

Thus, the individual focal lengths are:

- For the
**convex lens**: $ f_1 = 2f = 2 \times 5 = 10 $ cm. - For the
**concave lens**: $ f_2 = -3f = -3 \times 5 = -15 $ cm.

Therefore, the correct answer is:

**B: +10, -15**

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