Advertisements
Advertisements
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.
उत्तर
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Distance between the two lenses, d = 8.0 cm
(a) (i) When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
`1/"v"_1 - 1/"u"_1 = 1/"f"_1`
Where
`"u"_1` = Object distance = ∞
v1 = Image distance
`1/"v"_1 = 1/30- 1/∞ = 1/30`
∴ v1 = 30 cm
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
`1/"v"_2 - 1/"u"_2 = 1/"f"_2`
Where,
`"u"_2` = Object distance
= (30 − d) = 30 − 8 = 22 cm
`"v"_2` = Image distance
`1/"v"_2 = 1/22 - 1/20 = (10-11)/220 = (-1)/220`
∴ v2 = −220 cm
The parallel incident beam appears to diverge from a point that is `(220 - "d"/2 = 220 - 4)`216 m from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula, we have:
`1/"v"_2 - 1/"u"_2 = 1/"f"_2`
`1/"v"_2 = 1/"f"_2 + 1/"u"_2`
Where
`"u"_2` = Object distance = −∞
`"v"_2` = Image distance
`1/"v"_2 = 1/-20 + 1/-∞ = -1/20`
∴ v2 = −20 cm
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
`1/"v"_1 - 1/"u"_1 = 1/"f"_1`
Where,
`"u"_1` = Object distance
= −(20 + d) = −(20 + 8) = −28 cm
`1/"v"_1 = 1/30 + 1/-28 = (14 - 15)/420 = (-1)/420`
∴ v2 = −420 cm
Hence, the parallel incident beam appears to diverge from a point that is (420 − 4) 416 cm from the left of the centre of the combination of the two lenses.
The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.
(b) Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, `"u"_1` = −40 cm
`|"u"_1|` = 40 cm
According to the lens formula:
`1/"v"_1 - 1/"u"_1 = 1/"f"_1`
Where
`"v"_1` = Image distance
`1/"v"_1 = 1/30 + 1/(-40) = (4-3)/120 = 1/120`
∴ v1 = 120 cm
Magnification, `"m" = "v"_1/|"u"_1|`
= `120/40`
= 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula:
`1/"v"_2 - 1/"u"_2 = 1/"f"_2`
Where
`"u"_2` = Object distance
= +(120 − 8) = 112 cm
`"v"_2` = Image distance
`1/"v"_2 = 1/-20 + 1/112 = (-112 + 20)/2240 = (-92)/2240`
∴ v2 = `(-2240)/92` cm
Magnification, `"m'" = |"v"_2/"u"_2|`
= `2240/92 xx 1/112`
= `20/92`
Hence, the magnification due to the concave lens is `20/92`.
The magnification produced by the combination of the two lenses is calculated as: `"m" xx "m'"`
= `3 xx 20/92`
= `60/92`
= 0.652
The magnification of the combination is given as:
`"h"_2/"h"_1` = 0.652
h2 = 0.652 × h1
Where
h1 = Object size = 1.5 cm
h2 = Size of the image
∴ h2 = 0.652 × 1.5 = 0.98 cm
Hence, the height of the image is 0.98 cm.
APPEARS IN
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?
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
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?
An equiconvex lens of focal length 'f' is cut into two identical plane convex lenses. How will the power of each part be related to the focal length of the original lens ?
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 concave lenses L1 and L2 are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination
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.
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.
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 ______.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?
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.
Show that the least possible distance between an object and its real image in a convex lens is 4f, where f is the focal length of the lens.
