Physics · Wave Optics · NEET
Limit of resolution (smallest resolvable angle) d_theta = 1.22 lambda / D, in radians. Here lambda is the wavelength of light and D is the diameter (aperture) of the objective lens or mirror. Resolving power is the reciprocal: R.P. = D / (1.22 lambda). Two stars are 'just resolved' when their angular separation equals this d_theta.
A circular aperture (the round objective) produces a circular diffraction pattern called the Airy disc. Solving the diffraction of a plane wave through a circular hole gives the first dark ring at sin(theta) = 1.22 lambda / D. The 1.22 comes from the mathematics of a circular opening. For a rectangular slit the factor is just 1 (a sin theta = lambda), but a telescope's aperture is circular, so we use 1.22.
For a telescope, resolving power depends on the objective diameter D: R.P. = D / (1.22 lambda). We want the smallest ANGLE. For a microscope, resolving power depends on the numerical aperture: R.P. = 2 n sin(beta) / (1.22 lambda), and we want the smallest DISTANCE between two points. Telescope = far objects, angular resolution, bigger D is better. Microscope = near objects, linear resolution, larger n sin(beta) is better.
No. Magnification only makes the image bigger; it does not add new detail. Resolving power is fixed by the objective diameter D and the wavelength lambda. If two stars fall inside one Airy disc, they stay a single blur no matter how much you magnify (this is called empty magnification). To resolve them you need a bigger objective, not more magnification.
Because resolving power is directly proportional to D. Doubling the objective diameter halves the smallest resolvable angle, so the telescope can separate stars twice as close together. A larger D also gathers more light (brighter, fainter objects visible). This is why observatories use mirrors several metres wide.
Blue light has a smaller wavelength lambda. Since d_theta = 1.22 lambda / D, a smaller lambda gives a smaller (better) limit of resolution, so resolving power increases. Shorter wavelength = finer detail. That is why some instruments use blue or ultraviolet light for sharper images.
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 2 m is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the ability of a telescope to show two close stars as two separate points. Numerically, resolving power = D / (1.22 lambda), where D is the objective diameter and lambda is the wavelength. Larger D or smaller lambda gives higher resolving power.
It is the smallest angular separation between two objects that a telescope can just resolve. It equals 1.22 lambda / D radians. It is the reciprocal of resolving power - a smaller limit of resolution means a better (higher) resolving power.
The limit of resolution is an angle, so its unit is radian (rad). It has no other dimension because lambda and D are both lengths and their ratio is dimensionless.
Resolving power is directly proportional to the aperture (objective diameter D). Doubling D doubles the resolving power and halves the smallest resolvable angle, so more detail is seen. This is the main reason large telescopes have wide objectives.
No. Both use the 1.22 factor from circular-aperture diffraction, but a telescope's resolving power = D / (1.22 lambda) measures angular resolution, while a microscope's = 2 n sin(beta) / (1.22 lambda) measures linear resolution between nearby points.
