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.