Friday, February 09, 2007

Questions on Optical Instruments

The essential points you have to remember to work out questions on optical instruments are the following:
(1) Myopia (short sightedness) is corrected by a concave lens of focal length ‘d’ where ‘d’ is the distance of the far point in the case of the defective eye. [Note that for a normal eye the far point will be at infinity where as for the defective eye, it will be at a finite distance‘d’].
Therefore, f = –d
(2) Hypermetropia
(long sightedness) is corrected by a convex lens of focal length ‘f’ given by f = dD/(d – D) where ‘d’ is the distance of the near point for the defective eye and D is the least distance of distinct vision (25cm), [Note that for the normal eye, the near point will be at D and for the defective eye it will be at a greater distance ‘d’].
(3) Presbyopia is corrected by bifocal lens with the upper portion concave and the lower portion convex. [Note that for the defective eye in this case, the far point is nearer (and not at infinity) while the near point is farther away (and not at D)].
(4) Astigmatism is corrected by cylindrical lens.
(5) Magnifying power of a simple microscope M = 1+ D/f if the image is formed at the least distance of distinct vision ‘D’.
If the image is formed at infinity (normal setting or setting for relaxed eye), M= D/f .
(6) Magnifying power of compound microscope (M):
(i) M = (vo/uo)(1 + D/fe) if the final image produced by the eye piece is at the least distance of distinct vision ‘D’. Here uo is the distance of object from the objective, vo is the distance of the image produced by the objective and fe is the focal length of the eye piece.
(ii) If the final image is at infinity (normal setting or setting for relaxed eye),
M = (vo/uo)(1 + D/fe)
Approximate expressions
for the magnifying power of a compound microscope in the two cases are M = (L/fo)(1 + D/fe) and M = (L/fo)(D/fe) in the two cases respectively. Here L is the tube length of the microscope, which is the distance between the objective and the eye piece.
(7) (a)Limit of resolution (minimum separation between two point objects which can be resolved) of a microscope, dmin = λ /2n sinθ where λ is the wave length (in vacuum) of the light used for illuminating the object, ‘n’ is the refractive index of the medium between the object and the objective and θ is the semi angle of the cone of light proceeding from the object to the objective. This is Abbe’s expression for the resolving power when the object is not luminous and is therefore illuminated, as is usually the case. [Note that nsinθ is the numerical aperture]
(b) Resolving power of microscope = 1/dmin
(8) Magnifying power of telescope(M)
(i) M = β/α = fo/fe
for normal adjustment (image at infinity) where β is the angle subtended at the eye by the image, α is the angle subtended at the eye by the object, fo is the focal length of the objective and fe is the focal length of the eye piece.
(ii) M = (fo/fe)(1 + fe/D) if the final image is formed at the least distance of distinct vision ‘D’.
(9) Length of a telescope (in normal setting) = fo + fe
(10) Limit of resolution of telescope
(minimum angular separation between two point objects that can be resolved) dθ = 1.22λ/a where λ is the wave length of light proceeding from the object and ‘a’ is the aperture (diameter) of the objective.
(11) Resolving power of a telescope = 1/dθ = a/1.22λ
Now let us now consider the following MCQ:
The length of a microscope is 12 cm and its magnifying power is 25 for relaxed eye.The focal length of the eye piece is 4 cm. The distance of the object from the objective is
(a) 2 cm (b) 2.5 cm (c) 3 cm (d) 3.5 cm (e) 4 cm
The image produced by the objective is formed at the focus of the eye piece since the final image is formed at infinity (for relaxed eye). If vo is the distance of the image formed by the objective, the tube length of the microscope (distance between the objective and eye piece) is vo+ fe = vo+ 4 = 12 cm from which vo = 8 cm.
The magnifying power (for relaxed eye) is given by M = (vo/uo)(D/fe). Substituting the known values, 25 = (8/uo)(25/4) from which uo = 2 cm. [Option (a)].
The following question appeared in MPPMT 2000 question paper:
The length of the tube of a microscope is 10 cm. The focal lengths of the objective and eye lenses are 0.5 cm and 1 cm. The magnifying power of the microscope is about
(a) 5 (b) 166 (c) 23 (d) 500
The magnifying power, M = (L/fo)( D/fe) = (10/0.5)( 25/1) = 500.
Consider now the following MCQ which appeared in EAMCET 2000 question paper:
In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm. If an object is placed at 2 cm from the objective and the final image is formed at 25 cm from the eye lens, the distance between the two lenses is
(a) 6 cm (b) 7.75 cm (c) 9.25 cm (d) 11 cm
You should note that the focal length of the objective in a microscope is less than that of the eye piece. (In a telescope it is the other way round). Therefore, fo = 1.5 cm and fe = 6.25 cm.
The distance between the objective and the eye piece is the sum of the image distance(vo) for the objective and the object distance (ue) for the eye piece.
From the equation, 1/f = 1/v – 1/u as applied to the objective, we have
1/1.5 = 1/vo – 1/(–2).
Note that we have substituted the object distance as –2 in accordance with the Cartesian sign convention discussed in the post dated 22-11-06 (Questions (MCQ) on Refraction at Spherical Surfaces). This yields the image distance in the case of the objective as vo = 6 cm.The image distance for the eye piece is similarly given by
1/6.25 = 1/(–25) – 1/ue.
Note that the sign of the object distance is negative in accordance with the Cartesian sign convention. This equation yields ue = –5 cm. The negative sign just shows that the object distance for the eye piece is measured opposite to the direction of the incoming rays.In fact, we know that the real image (of the object) formed by the objective serves as the object for the eye piece and therefore its distance is negative. However, since ue is an unknown quantity, we did not bother about its sign. If we had substituted its sign as negative in the law of distances, we would have obtained the value as +5 cm.
The distance between the objective and the eye piece is 6+5 =11 cm.
Here is a simple question involving myopia:
A man who cannot see clearly beyond 10m wants to see stars clearly. He should use a lens of power (in dioptre)
(a) 10 (b) – 10 (c) –1 (d) 0.1 (e) – 0.1
His defect is myopia and hence he should use a concave lens of focal length equal to the distance of the far point, which is given as 10m. The power of the lens is 1/(–10) = – 0.1 dioptre.
Now consider the following question:
The near point of a person with defective eye is at 65 cm. To correct his defect, he should use spectacle lenses of focal length
(a) 65 cm (b) 55.6 cm (c) 50.6 cm (d) 45,5 cm (e) 40.6 cm
As his defect is hypermetropia (long sightedness), he should use convex lenses of focal length ‘f’ given by f = dD/(d – D) where ‘d’ is the distance of his far point and ‘D’ is the least distance of distinct vision. Therefore, f = 65×25/(65 – 25) = 40.6 cm.
Here is a typical simple question on telescopes:
The magnifying power of a small telescope is 25 and the separation between its objective and eye piece is 52 cm in normal setting. The focal lengths of its objective and eye piece are respectively
(a) 50 cm and 2 cm (b) 27 cm and 25 cm (c) 45 cm and 7 cm (d) 50.5 cm and 1.5 cm (e) 47 cm and 5 cm
We have magnifying power, M = fo/fe so that 25 = fo/fe, from which fo = 25 fe.
Sincethe length of the telescope = fo + fe, we have 52 = 25fe + fe = 26 fe so that fe = 2 cm and hence fo = 50 cm.
The following question pertains to the resolving power of a telescope:
The distance between the earth and the moon is nearly 3.8×105 km. What is the separation of two points on the moon that can be just resolved using a 400 cm telescope, using light of wave length 6000 Ǻ?
(a) 46.5 m (b) 56 m (c) 69.5 m (d) 78.6 m (e) 85.5 m
The limit of resolution dθ = 1.22λ/a. This is the minimum angular separation between objects that can be just resolved.
The linear separation between the objects that can be just resolved is rdθ = 3.8×108×1.22×6000×10–10/4 = 69.5 m. [Note that we have converted all distances in to metre].
The following m.c.q. appeared in the MPPMT 2000 question paper:
The focal lengths of the eye piece and objective of a telescope are respectively 100 cm and 2 cm. The moon subtends an angle of 0.5º at the eye. If it is looked through the telescope, the angle subtended by the moon’s image will be
(a) 100º (b) 25º (c) 50º (d) 10º
the magnifying power of a telescope is given by M = β/α = fo/fe with usual notations. Therefore, β = α fo/fe = 0.5×100/2 = 25º [Option (b)].
Now Consider the following MCQ:
A telescope has an objective lens of focal length 1.5m and an eye piece of focal length 5 cm. If this telescope is used in normal setting to view a tower of height 100m located 3km away, what will be the height of the image of the tower?
(a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm (e) 50 cm
The angle subtended at the objective by the tower will be the same as the angle subtended (at the objective) by the image produced by the objective so that we have
100/3000 = h/1.5 where ‘h’ is the height of the image produced by the objective. [Note that this image is at the focus of the objective].
From the above, h = 0.05m = 5 cm.
The magnifying power of the eye piece in normal adjustment is D/fe = 25cm/5cm = 5.
The height of the final image = 5h = 25 cm.

2 comments:

  1. Hi MK, Nice to see your sites on Physics and entrance. I am a BSc Physics graduate myself and I think this site would be really useful to people.

    One recommendation I have is that it may be easier if you break down some of your posts into smaller posts. It may be helpful for your readers.

    BTW, thanks for visiting my blog.

    ReplyDelete
  2. Thank you Krishnakumar, for your comments. Your suggestion is definitely welcome.
    With best wishes
    VM

    ReplyDelete