Question

Focal length of a thin convex lens placed in air is 10 cm. An object is placed at a distance of 5 cm from the first focus. The distance of the image from the second focus is :

• A. 20 cm
• B. 15 cm
• C. 30 cm
• D. 25 cm

Hint:

For a thin convex lens, the total distance between the object and the lens is the sum of the distances from the first focus to the lens and from the lens to the object.

Answer 1

The Correct Option is B

Step 1: Understanding the Focal Length and Focus of a Lens.
The focal length \( f \) of a convex lens is the distance from the lens to its focus. The first and second foci of the lens are located symmetrically on either side of the lens. Here, the given focal length of the lens is 10 cm.
Step 2: Object Placement and Image Formation.
The object is placed at a distance of 5 cm from the first focus. The total distance between the object and the lens is the sum of the distance from the first focus to the lens and the distance from the lens to the object: \[ \text{Distance from lens to object} = 5 \, \text{cm} + 10 \, \text{cm} = 15 \, \text{cm} \] Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Since the object distance \( u = -15 \, \text{cm} \) (negative because the object is on the same side as the incoming light), and \( f = 10 \, \text{cm} \), we can substitute into the formula: \[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \] Solving for \( v \): \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \] \[ v = 30 \, \text{cm} \]
Step 3: Conclusion.
Thus, the distance of the image from the second focus is 15 cm, corresponding to option .