Advertisements
Advertisements
Question
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.
Advertisements
Solution
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)
RELATED QUESTIONS
- 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.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
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 pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.
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.
The point near the centre of a thin lens through which a ray of light passes without appreciable deviation is called the:
Which of the following correctly represents the Lens Maker's Formula?
According to the new Cartesian sign convention, all distances in lens problems are measured from:
Using the sign convention with \[OB = -u\] and \[DI = +v\] in the combined refraction equation for a thin lens gives:
Magnification produced by a lens is positive when the image is:
