Question

(a) Upper half of a convex lens is covered with a black paper. Draw a ray diagram to show the formation of image of an object placed at a distance of 2F from such a lens. Mention the position and nature of the image formed. State the observable difference in the image obtained if the lens is uncovered. Give reason to justify your answer.
(b) An object is placed at a distance of 30 cm from the optical centre of a concave lens of focal length 15 cm. Use lens formula to determine the distance of the image from the optical centre of the lens.

Answer 1

(a) Image formation by a convex lens with upper half covered:
When the upper half of a convex lens is covered with black paper, only the lower half of the lens allows light to pass. However, the lens still forms a full image because all parts of the lens contribute to forming the same complete image.

Ray diagram description:
- Place the object at 2F (i.e., twice the focal length) on one side of the convex lens.
- A ray parallel to the principal axis from the object passes through the lower half of the lens and refracts through the focus on the other side.
- Another ray passing through the optical centre goes straight without deviation.
- These rays meet at the other side at a point beyond F and 2F — forming the image.

Position of image: At 2F on the other side of the lens
Nature of image: Real, inverted, same size as the object

Observable difference when lens is uncovered:
- The position, size, and nature of the image remain the same.
- The brightness of the image is reduced because less light passes through the lens.

Reason: Only the amount of light (intensity) is affected when part of the lens is blocked. The remaining part still bends light rays correctly to form a complete image.

(b) Use of lens formula to find image distance for a concave lens
Given:
- Object distance (u) = -30 cm (object is on the left of the lens, hence negative)
- Focal length (f) = -15 cm (for concave lens, focal length is negative)

Lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substitute the values:
\[ \frac{1}{-15} = \frac{1}{v} - \frac{1}{-30} \]
\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{30} \]
Take LCM and solve:
\[ \frac{1}{v} = \frac{1}{-15} - \frac{1}{30} = \frac{-2 - 1}{30} = \frac{-3}{30} = \frac{-1}{10} \]
So, \[ v = -10 \text{ cm} \]

Final Answer:
- Distance of image: 10 cm from the optical centre of the lens on the same side as the object (virtual and upright).