A lens having focal length f and aperture of diameter d forms an image...
Focal length of the lens remains same.
Intensity of image formed by lens is proportional to area exposed to incident light from object.
i.e. Intensity ∝ area
After blocking, exposed area,
Hence, focal length of a lens = f, intensity of the image = 3I/4
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A lens having focal length f and aperture of diameter d forms an image...
Given:
- Lens with focal length f and aperture diameter d forms an image of intensity I.
- Aperture of diameter d/2 in the central region of the lens is covered by a black paper.
To find:
- The new focal length of the lens.
- The new intensity of the image.
Solution:
When the aperture is covered by black paper, it acts as a stop, limiting the amount of light passing through the lens. This causes a decrease in the intensity of the image formed.
1. Effect of covering the aperture:
- When the aperture is covered, the effective diameter of the lens is reduced to d/2.
- The area of the effective aperture is proportional to the square of its diameter. So, the area of the effective aperture is reduced to (d/2)^2 = d^2/4.
- Therefore, the amount of light passing through the lens is reduced by a factor of 1/4.
2. Effect on focal length:
- The focal length of a lens is determined by its shape and refractive index, and it does not depend on the diameter of the aperture.
- Therefore, covering the aperture does not change the focal length of the lens. Hence, the new focal length of the lens remains f.
3. Effect on intensity:
- The intensity of the image is given by the formula I ∝ (1/f^2).
- Since the focal length remains unchanged, the intensity of the image is directly proportional to 1/f^2. Therefore, the intensity of the image remains I.
Hence, the new focal length of the lens is f, and the new intensity of the image is 3I/4. Therefore, the correct option is (c) f and 3I/4.
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