Question

One convex lens and one concave lens placed is contact with each other. If the ratio of their power is \((2 / 3)\) and focal length of the combination is \(30 \mathrm{~cm}\), then individual focal lengths are (A) \(15 \mathrm{~cm}\) and \(-10 \mathrm{~cm}\) (B) \(-15 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) (C) \(30 \mathrm{~cm}\) and \(-20 \mathrm{~cm}\) (D) \(-30 \mathrm{~cm}\) and \(-30 \mathrm{~cm}\)

Short answer

The individual focal lengths of the lenses are \(15~cm\) for the convex lens and \(-10~cm\) for the concave lens. The correct option is (A).

Step-by-step solution

Understand the lens powers and their relationship with the focal lengths

Lens power is the reciprocal of its focal length. For a convex (converging) lens, the power is positive, while for a concave (diverging) lens, the power is negative. In the exercise, we are given the ratio of their powers as \(2/3\), and we can use this relationship to relate their focal lengths: Let the focal length of the convex lens be \(f_1\) and the focal length of the concave lens be \(f_2\). According to the given ratio of powers, we have: \[\frac{1/f_1}{1/f_2} = \frac{2}{3}\]

Combine the powers of lenses to find the focal length of the combination

When two lenses are placed in contact, their combined power is the sum of their individual powers. Hence, their combined focal length can be found using the lens formula: \[\frac{1}{F_{combination}}=\frac{1}{f_1}+\frac{1}{f_2}\] We are given the focal length of the combination as \(30~cm\). Substitute the value of the focal length of combination in the equation: \[\frac{1}{30}= \frac{1}{f_1}+\frac{1}{f_2}\]

Solve the equations for individual focal lengths

Now, we have two equations with the two unknowns \(f_1\) and \(f_2\). To solve for these focal lengths, we can use the method of substitution or elimination. Equation 1: \[\frac{1/f_1}{1/f_2}=\frac{2}{3}\] Equation 2: \[\frac{1}{30}=\frac{1}{f_1}+\frac{1}{f_2}\] From equation 1, we can write: \[3f_2= 2f_1\] Now, substitute \(f_2\) in terms of \(f_1\) in equation 2: \[\frac{1}{30}=\frac{1}{f_1}+\frac{1}{(3/2)f_1}\] Now, solve for \(f_1\): \[f_1 = 15~cm\] Now, use the relation \(3f_2 = 2f_1\) to find \(f_2\): \[f_2 = -10~cm\]

Interpret the results

The individual focal lengths of the lenses are found to be \(15~cm\) for the convex lens and \(-10~cm\) for the concave lens. This corresponds to option (A) in the given choices.

Key concepts

Convex Lens

A convex lens, often referred to as a converging lens, is a transparent optical device that bends light rays inward, focusing them at a point on the opposite side of the lens. These lenses have a bulging shape that is thicker in the middle than at the edges, similar to an oval or an almond.
Convex lenses are crucial in many optical instruments such as cameras, glasses for farsightedness, and telescopes. They play a key role in magnifying images and adjusting the focus in various optical systems.

• The focal length of a convex lens is the distance from the lens where parallel rays of light converge to a single point.
• This focal length is positive, which reflects the converging nature of the lens.

In the original exercise, the focal length of the convex lens ( f_1 ) is determined to be 15 cm, indicating its ability to bend light rays to meet at a focal point 15 centimeters away from the lens.

Concave Lens

A concave lens, or diverging lens, is an optical device designed to spread out light rays, making it appear as though they originate from a single point. These lenses have a thinner middle and thicker edges, resembling a bow tie shape.
Concave lenses are used in a variety of applications, including correcting nearsightedness in eyeglasses, peepholes in doors, and optimizing beam expanders.

• A concave lens has a negative focal length, reflecting its diverging characteristic.
• The negative sign indicates that it disperses light rather than converging it.

In the context of the original exercise, the focal length of the concave lens ( f_2 ) was computed to be -10 cm.
This negative value demonstrates how the lens diverges light rays, causing them to appear as though they are coming from a virtual focal point located 10 centimeters on the same side of the lens as the incoming light.

Lens Power

The power of a lens quantifies its ability to bend light, bringing it into focus. It is defined as the reciprocal of its focal length, usually measured in diopters (D).
A high power means a lens sharply bends light, while low power indicates a gentler bend.

• Convex lenses have positive power due to their converging action.
• In contrast, concave lenses have negative power reflecting their divergent effect on light.

The original exercise provides a ratio of lens powers as \( \frac{2}{3} \). This means the power of the convex lens is two-thirds of that of the concave lens.
This relationship helps in solving for the individual focal lengths since the power is inversely related to focal length, impacting how each lens affects light.

Focal Length Combination

When lenses are placed in contact, forming a lens system, their combined power creates a new focal length, called the focal length of the combination. This combination alters how the lenses work together, affecting the path and focus of light rays.
The equation \[ \frac{1}{F_{\text{combination}}} = \frac{1}{f_1} + \frac{1}{f_2} \] expresses this relationship. Here, \( F_{\text{combination}} \) is the focal length of the combined lenses, \( f_1 \) and \( f_2 \) are the focal lengths of the individual lenses.

• This formula is critical for systems using multiple lenses, such as microscopes and telescopes, providing a straightforward method to evaluate their optical power.
• In the exercise, the combined focal length is specified as 30 cm, guiding us to determine the individual characteristics of each lens.

The lens system, in this case, results in focal lengths of 15 cm for the convex lens and -10 cm for the concave lens, highlighting how individual properties impact the overall function of the system.