Question

The distance of an object from the first focal point of a biconvex lens is \( X_1 \) and distance of the image from second focal point is \( X_2 \). The focal length of the lens is:

• A. \( X_1 X_2 \)
• B. \( \sqrt{X_1 + X_2} \)
• C. \( \sqrt{X_1 X_2} \)
• D. \( \frac{X_2}{X_1} \)

Hint:

For a biconvex lens, the focal length can be found using the relationship \( f = \sqrt{X_1 X_2} \), where \( X_1 \) and \( X_2 \) are the distances from the focal points.

Answer 1

The Correct Option is C

In the given problem, we aim to determine the focal length \( f \) of a biconvex lens based on the distances from the focal points. We start by using the lensmaker's formula for a biconvex lens:

Step 1: Understand the lensmaker's formula, which relates object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)) as:

\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]

However, given that the object is at distance \( X_1 \) from the first focal point, the effective object distance \( u' \) becomes:

\[ u = X_1 + f \]

Similarly, the image is at distance \( X_2 \) from the second focal point, making the effective image distance \( v' \) as:

\[ v = X_2 + f \]

Step 2: Substitute these effective distances in the lens formula:

\[\frac{1}{f} = \frac{1}{X_1 + f} + \frac{1}{X_2 + f}\]

Step 3: Simplifying and considering that this forms a special case where the lens is very thin (lensmaker's equation symmetry condition), leads to:

\[\Rightarrow X_1 X_2 = f^2\]

Step 4: Solving for the focal length \( f \):

\[ f = \sqrt{X_1 X_2} \]

Thus, the focal length of the lens is \(\sqrt{X_1 X_2}\).