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 Obscured:
When the upper half of a convex lens is covered, only the lower half transmits light. Nevertheless, a complete image is still formed because all sections of the lens contribute to creating the entire image.

Ray Diagram Explanation:
- Position the object at 2F on one side of the convex lens.
- A ray from the object, parallel to the principal axis, travels through the lower half of the lens and refracts through the focal point on the opposite side.
- A second ray, passing through the optical center, continues undeviated.
- These refracted rays converge beyond F and 2F on the opposite side, thereby forming the image.

Image Location: At 2F on the side opposite the object.
Image Characteristics: Real, inverted, and the same size as the object.

Observable Changes When Lens is Uncovered:
- The image's position, size, and nature remain unchanged.
- The image brightness diminishes due to reduced light transmission through the lens.

Rationale: Obscuring part of the lens only affects the quantity of light (intensity). The unobstructed portion of the lens continues to refract light rays correctly to form a complete image.

(b) Determining Image Distance for a Concave Lens Using the Lens Formula
Given:
- Object distance (u) = -30 cm (object is to the left of the lens, hence negative)
- Focal length (f) = -15 cm (focal length is negative for a concave lens)

Lens Formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the given values:
\[ \frac{1}{-15} = \frac{1}{v} - \frac{1}{-30} \]
\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{30} \]
Calculating the common denominator and solving for \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{-15} - \frac{1}{30} = \frac{-2 - 1}{30} = \frac{-3}{30} = \frac{-1}{10} \]
Therefore, \[v = -10 \text{ cm}\]

Conclusion:
- Image Distance: 10 cm from the lens's optical center, on the same side as the object (resulting in a virtual and upright image).