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) • When an object is placed at a distance of 2F from a convex lens, the image is formed at 2F on the other side of the lens.
• The image is real, inverted, and of the same size as the object.
• If the upper half of the lens is covered with black paper, the entire image is still formed, but its brightness is reduced.
• If the lens is uncovered, the image formed will be of normal brightness.
• Reason: The uncovered half of the lens refracts light, so the entire image is formed, albeit less bright.
Ray Diagram:

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(b) The lens formula is:
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u},\]
where:
\( f = -15 \, \text{cm} \) (negative for concave lens),
\( u = -30 \, \text{cm} \) (negative because the object is on the same side as the light source),
\( v = ? \) (distance of the image).
Substitute the values into the formula:
\[\frac{1}{-15} = \frac{1}{v} - \frac{1}{-30}.\]
Simplify:
\[\frac{1}{-15} = \frac{1}{v} + \frac{1}{30}.\]
Take the LCM of 15 and 30:
\[\frac{1}{-15} = \frac{1}{v} + \frac{2}{60}.\]
\[\frac{1}{-15} - \frac{1}{30} = \frac{1}{v}.\]
Simplify further:
\[\frac{-2 + 1}{30} = \frac{1}{v}.\]
\[\frac{-1}{30} = \frac{1}{v}.\]
\[v = -30 \, \text{cm}.\]
Thus, the image is formed 30 cm on the same side as the object (negative sign indicates the same side).
Nature of the Image:
The image is virtual, erect, and diminished.