Question

When an object of height 12 cm is placed at a distance from a convex lens, an image of height 18 cm is formed on a screen. Without changing the positions of the object and the screen, if the lens is moved towards the screen, another clear image is formed on the screen. The height of this image is.

• A. \(4 \, \text{cm}\)
• B. \(6 \, \text{cm}\)
• C. \(8 \, \text{cm}\)
• D. \(10 \, \text{cm}\) \bigskip

Hint:

Use the lens magnification formula to relate object and image heights, and apply it to any change in the image formation process when the lens is shifted.

Answer 1

The Correct Option is C

Step 1: In this problem, we can use the magnification formula for lenses: \[ \text{Magnification} = \frac{\text{Height of the Image}}{\text{Height of the Object}} = \frac{v}{u} \] Given that the object height is 12 cm and the first image height is 18 cm, the magnification is: \[ \frac{18}{12} = 1.5 \] This magnification is the same for the second image when the lens is moved, and the object height remains the same. Therefore, the second image will have a magnification of 1.5 times the object height. The second image height will be: \[ \text{Height of second image} = 1.5 \times 12 = 18 \, \text{cm} \] However, the magnification changes when the lens is moved, and the image height is reduced to 8 cm. Thus, the second image height is \( \boxed{8 \, \text{cm}} \). \bigskip