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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
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
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(a) the image will be shifted downward
(b) the image will be shifted upward
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(d) the intensity of the image will decrease.
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Answer the following question.
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(a) Identify the optical instrument.
(b) Calculate the magnification produced by the instrument.
Will the focal length of a lens for red light be more, same or less than that for blue light?
In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.
