Advertisements
Advertisements
Question
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Advertisements
Solution
Focal length of the objective lens, fo = 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
`"m"_"e" = (1 + "d"/"f"_"e")`
= `(1 + 25/5)`
= 6
The angular magnification of the objective lens (mo) is related to me as:
mo me = m
mo = `"m"/"m"_"e"`
= `30/6`
= 5
We also have the relation:
`"m"_"o" = ("Image distance for the objective lens" ("v"_"o"))/("Object distance for the objective lens" (-"u"_"o"))`
`5 = "v"_"o"/-"u"_"o"`
∴ vo = −5uo ................(1)
Applying the lens formula for the objective lens:
`1/"f"_"o" = 1/"v"_"o" - 1/"u"_"o"`
`1/1.25 = 1/(-5"u"_"o") - 1/"u"_"o"`
`1/1.25 = (-6)/(5"u"_"o")`
uo = `(-6)/5 xx 1.25`
∴ uo = −1.5 cm
And vo = −5uo
= −5 × (−1.5)
vo = 7.5 cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
`1/"v"_"e" - 1/"u"_"e" = 1/"f"_"e"`
Where,
ve = Image distance for the eyepiece = −d = −25 cm
ue = Object distance for the eyepiece
`1/"u"_"e" = 1/"v"_"e" - 1/"f"_"e"`
= `(-1)/25 - 1/5`
= `- 6/25`
∴ ue = −4.17 cm
Separation between the objective lens and the eyepiece = `|"u"_"e"| + |"v"_"o"|`
= 4.17 + 7.5
= 11.67
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
APPEARS IN
RELATED QUESTIONS
Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision
When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
When are two objects just resolved? Explain.
Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can be corrected.
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.
Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point.
In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope.
In which of the following the final image is erect?
(a) Simple microscope
(b) Compound microscope
(c) Astronomical telescope
(d) Galilean telescope
A simple microscope using a single lens often shows coloured image of a white source. Why?
The magnifying power of a converging lens used as a simple microscope is `(1+D/f).` A compound microscope is a combination of two such converging lenses. Why don't we have magnifying power `(1+D/f_0)(1+D/f_0)`?In other words, why can the objective not be treated as a simple microscope but the eyepiece can?
A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?
compound microscope consists of two convex lenses of focal length 2 cm and 5 cm. When an object is kept at a distance of 2.1 cm from the objective, a virtual and magnified image is fonned 25 cm from the eye piece. Calculate the magnifying power of the microscope.
Draw a labelled ray diagram showing the formation of image by a compound microscope in normal adjustment. Derive the expression for its magnifying power.
On increasing the focal length of the objective, the magnifying power ______.
Draw a ray diagram of compound microscope for the final image formed at least distance of distinct vision?
An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope for the final image formed at least distance of distinct vision?
| A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length, is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye-piece, in turn, produces the final magnified image. |
Which of the following is not correct in the context of a compound microscope?
| A compound microscope consists of two converging lenses. One of them, of smaller aperture and smaller focal length, is called objective and the other of slightly larger aperture and slightly larger focal length is called eye-piece. Both lenses are fitted in a tube with an arrangement to vary the distance between them. A tiny object is placed in front of the objective at a distance slightly greater than its focal length. The objective produces the image of the object which acts as an object for the eye-piece. The eye-piece, in turn, produces the final magnified image. |
A compound microscope consists of an objective of 10X and an eye-piece of 20X. The magnification due to the microscope would be:
