Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
From the reversibility of u and v, as seen from the formula for lens,
`1/f = 1/v - 1/u`
It is clear that there are two positions for which there shall be an image on the screen.
Let the first position be when the lens is at O.
Given `-u + v = D`
⇒ `u = -(D - v)`
Placing it in the lens formula
`1/(D - v) + 1/v = 1/f`
⇒ `(v + D - v)/((D - v)v) = 1/f`
⇒ `v^2 - Dv + Df` = 0
⇒ `v^2 = D/2 +- sqrt(D^2 - 4Df)/2`
`u = -(D - v) = - (D/2 + sqrt(D^2 - 4Df)/2)`
Thus, if the object distance is `D/2 - sqrt(D^2 - 4Df)/2` then the image is at `D/2 + sqrt(D^2 - 4Df)/2`
If the object distance is `D/2 + sqrt(D^2 - 4Df)/2`, then the image is at `D/2 - sqrt(D^2 - 4Df)/2`.
The distance between the poles for these two object distances is
`D/2 - sqrt(D^2 - 4Df)/2 - (D/2 - sqrt(D^2 - 4Df)/2) = sqrt(D^2 - 4Df)`
Let `d = sqrt(D^2 - 4Df)`
If `u = D/2 + d/2` then the image is at `v = D/2 - d/2`
∴ The magnification `m_1 = (D - d)/(D + d)`
If `u = (D - d)/2` then `v = (D + d)/2`
∴ The magnification `m_2 = (D + d)/(D - d)` Thus `m_2/m_1 = ((D + d)/(D - d))^2`.
APPEARS IN
संबंधित प्रश्न
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
- a convex lens of focal length 20 cm, and
- a concave lens of focal length 16 cm?
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 small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sum is just setting? Refractive index of water = 4/3.
Answer the following question.
An optical instrument uses a lens of 100 D for the objective lens and 50 D for its eyepiece. When the tube length is kept at 20 cm, the final image is formed at infinity.
(a) Identify the optical instrument.
(b) Calculate the magnification produced by the instrument.
An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image ______.
Will the focal length of a lens for red light be more, same or less than that for blue light?
A plano convex lens has diameter of 10 cm and its thickness at the centre is 0.5 cm. Speed of light in the lens is 2 × 108 ms-1. What is the focal length of the lens?
A transparent refracting medium bounded by two surfaces, of which at least one is spherical, is called a:
The point on the principal axis where rays parallel to the principal axis actually meet after refraction, or appear to diverge after refraction, is called the:
The distance between the optical centre and the principal focus is called the:
According to the new Cartesian sign convention used for lenses, distances measured opposite to the direction of incident light are taken as:
According to the new Cartesian sign convention for lenses, heights measured below the principal axis are taken as:
According to the new Cartesian sign convention, all distances in lens problems are measured from:
A ray directed towards the principal focus of a concave lens, after refraction, emerges:
Using the sign convention with \[OB = -u\] and \[DI = +v\] in the combined refraction equation for a thin lens gives:
A lens with an infinite focal length behaves like:
Magnification produced by a lens is positive when the image is:
