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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.
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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.
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?
You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
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?
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(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
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.
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 ______.
A transparent refracting medium bounded by two surfaces, of which at least one is spherical, is called a:
The point near the centre of a thin lens through which a ray of light passes without appreciable deviation is called the:
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 ratio of the height of the image to the height of the object is called:
Which of the following correctly represents the Lens Maker's Formula?
According to the new Cartesian sign convention for lenses, the focal length of a convex lens is:
According to the new Cartesian sign convention, all distances in lens problems are measured from:
A lens with an infinite focal length behaves like:
Magnification produced by a lens is positive when the image is:
