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प्रश्न
In the given figure the radius of curvature of the curved face in the planoconvex and the planoconcave lens is 15 cm each. The refractive index of the material of the lenses is 1.5. Find the final position of the image formed.
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उत्तर
The focal length of the plano-convex lens is:
`1/f = (1.5 - 1)(1/((+10)) - 1/∞) = 1/20`
`f = 20`
The focal length of plano-concave lens is:
`1/f = (1.5 - 1)(1/∞ - 1/10) = -1/20`
`f = -20`
Since parallel beams are incident on the lens, its image from plano-concave lens will be formed at +30 cm from it (at the focus) and will act as an object for plano-concave lens. Since the two lenses are separated by a distance of 20 cm, therefore for the next lens.
u = +10 cm
`v = (uf)/((u + f)) = (10 xx(- 20))/(10 + (- 20)`
`v = (-200)/(-10)`
`v` = 20 cm (final position of the image formed)
संबंधित प्रश्न
- Determine the ‘effective focal length’ of the combination of the two lenses, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of the effective focal length of this system useful at all?
- An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system and the size of the image.
A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
- What is the magnification produced by the lens? How much is the area of each square in the virtual image?
- What is the angular magnification (magnifying power) of the lens?
- Is the magnification in (a) equal to the magnifying power in (b)? Explain.
A double convex lens of + 5 D is made of glass of refractive index 1.55 with both faces of equal radii of curvature. Find the value of its radius of curvature.
Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without loosing the energy of the light. This increase the intensity. Describe how the converging lenses should be placed to do this.
A convex lens forms a real image of a point object placed on its principals axis. If the upper half of the lens is painted black,
(a) the image will be shifted downward
(b) the image will be shifted upward
(c) the image will not be shifted
(d) the intensity of the image will decrease.
For a thin lens, the magnification \[m\] in terms of image distance \[v\] and object distance \[u\] is given by:
According to the new Cartesian sign convention for lenses, the focal length of a concave lens is:
A ray parallel to the principal axis, after refraction through a concave lens, appears to diverge from:
A ray passing through the principal focus of a convex lens, after refraction, emerges:
If the refractive index of a liquid equals that of a glass lens (\[n_1 = n_2\]), the lens maker relation gives \[\frac{1}{f} = 0\]. This means the focal length of the lens becomes:
