Two Questions (MCQ) on Telescopes

nvppfThe following question appeared in MPPMT 2000 question paper:
The diameter of the objective of a telescope is ‘a’, its magnifying power is ‘m’ and the wave length of light is λ. The resolving power of the telescope is
(a) a/1.22 λ (b) λm/1.22a (c) 1.22a/ λ (d) a/1.22 λm
The magnifying power given in the problem is just a distraction. The limit of resolution of a telescope, which is the minimum angular separation between two objects that can be resolved is given by dθ = 1.22λ/a. The resolving power is the reciprocal of the limit of resolution and is equal to a/1.22 λ. [Option (a)].
Objectives and eye pieces of telescopes are made of achromatic combination of lenses to make them devoid of the defect of chromatic aberration. Consider the following MCQ:
A telescope objective is an achromatic doublet made of crown glass and flint glass lenses. The proper choice for the achromatic combination is
(a) divergent lens of flint glass and convergent lens of crown glass
(b) convergent lens of flint glass and divergent lens of crown glass
(c) both convergent (d) both divergent
(e) as given in options (a) or (b) or (c)
You might be remembering the condition for an achromatic doublet: ω₁ /f₁+ ω₂/f₂ = 0, from which ω₁/f₁= – ω₂/f₂.
Since the dispersive power ‘ω’ is a positive quantity, the negative sign shows that one lens must be diverging (negative focal length) and the other converging. In a telescope, the objective has to be converging. Therefore, the converging lens should have smaller focal length (larger power) so that when combined with the diverging lens, the combination will still be converging. Crown glass has smaller dispersive power and hence the converging lens should be made of crown glass to satisfy the above condition of achromatism [Option (a)].

You will find more multiple choice questions (with solution) on telescopes, microscopes and the defects of the eye at physicsplus: Questions on Optical Instruments

Optics - Questions on Reflection and Refraction

The average student is usually scared of questions in ray optics because of the wrong notion that the sign convention is some thing that adds to one’s confusion! The best thing to do is to work out as many questions as possible. You will definitely realize that there is nothing to worry about. Let us discuss the following questions:
(1) A glass slab of refractive index 1.5 and thickness 4.5cm is kept close to a concave mirror of focal length 20cm so that the face of the slab is perpendicular to the principal axis of the mirror. A point object is placed on the principal axis so that its image coincides with itself. Then, the distance of the object from the mirror is
(a) 40cm (b) 41.5cm (c) 43cm (d) 37cm (e) 47.5cm
If the glass slab is not interposed, the object is to be placed at the center of curvature of the mirror (at 40cm). When the glass slab is present, the distance of the object is to be increased by the apparent shift t - (t/μ) produced by the slab, so that the ray emerging from the slab appears to come from the center of curvature of the mirror.
Therefore, distance of the object = 40+ 4.5 – (4. 5/1. 5) = 41.5cm [option (b)].

(2) A glass sphere of radius 2cm and refractive index 1.5 has a small air bubble at a distance of 1.2cm from the surface through which it is viewed normally. Its apparent distance from the surface is
(a) 0.8cm (b) 1cm (c) 1.33cm (d) 1.6cm (e) 2.4cm
Don’t be tempted to pick out option (a), under the impression that this is similar to the normal refraction at a glass slab. You have to use the equation, μ₂/v - μ₁/u = (μ₂- μ₁)/R where μ₂= 1, μ₁= 1.5, R = -2cm and u = -1.2cm.
Note that we have imagined the ray from the object (air bubble) to be proceeding in the positive X-direction so that the object distance measured from the pole (which is imagined to be at the origin) to the object is negative. The radius of curvature of the spherical surface also is negative since we have to measure it from this pole to the centre of curvature. We leave the unknown quantity ‘v’ in the equation as it is, with out worrying about its sign.
Substituting these values, v = -1cm. The negative sign shows that the image, which is virtual, is on the same side of the refracting surface as the object is.
(3)An astronomical telescope has an eye piece of focal length 10cm. It gives a magnification of 20 in normal adjustment. The distance between the objective and the eye piece is
(a) 30cm (b) 60cm (c) 110cm (d) 200cm
In normal adjustment, magnification, m = fo/fe where fo and fe are the focal lengths of the objective and eye piece respectively. Therefore, fo = m.fe = 20´10 = 200cm.
Distance between objective and eye piece = fo + fe = 200+10 = 210cm.
(4) The plane faces of two identical plano-convex lenses each having a focal length of 50cm are placed against each other to form a usual biconvex lens. The distance from this lens combination at which an object must be placed to obtain a real, inverted image which has the same size as the object is
(a) 50cm (b) 25cm (c) 100cm (d) 40cm (e) 125cm
This simple question appeared in Kerala Engineering Entrance test paper of 2006. When two identical lenses are kept in contact, the focal length of the combination is reduced to half the value of each lens. (This is true even if the plane surface of one plano-convex lens is placed in contact with the curved surface of the other). The focal length of the combination therefore is 25cm.
A converging lens will produce a real, inverted image of the same size as the object when the object distance is 2f. So, the answer is 50cm.

You will find more multiple choice questions (with solution) from Optics here as well as at physicsplus: Questions (MCQ) on Refraction at Spherical Surfaces