Define resolving power of an astronomical refracting telescope and write expression for it in normal adjustment. Assume that light of wave length 6000Å is coming from a star, what is the limit of resolution of a telescope whose objective has a diameter of 2.54m?

OR

Write the basic assumptions used in the derivation of lens – maker’s formula and hence derive this expression.

The resolving power of the telescope is defined as the ability of the instrument to produce separate images of two closely spaced objects.

Mathematically, it can be represented as,

Where D is the diameter or the aperture of objective lens and λ is the wavelength of the light falling on the objective lens.

Now, it is given in the question, diameter of the objective = 2.54 inch

Wavelength λ= 6000 Å = 6 x 10^-10 m

Resolving power =

OR

Lens-maker’s formula provides the relation between the focal length of the lens, the refractive index of the its material and the radii of curvature of its two surfaces.

Assumptions made during the derivation of the formula are:

a) The lens is thin so that the distance measured from the poles of the two surfaces of the lens can be taken to be equal to the distances measured from the optical centre.

b) The object is a point object which is placed on the principal axis.

c) The aperture of the lens is small.

d) The incident and refracted ray makes small angle with the principal axis.

Derivation:

∆ A’B’O and ∆ ABO are similar

Also, ∆ A’B’F and ∆ MOF are similar,

But we know from the ray diagram, MO=AB,

Using equations (1) and (2), we can get

According to the sign convention,

BO= object distance= -u

OB’= image distance = +v

OF= focal length = +f

Therefore,

Divide the whole equation by uvf, we get