Question

(i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain briefly the working. (ii) An astronomical telescope uses two lenses of power 10 D and 1 D the what is the magnifying power in normal adjustment?​

Answer 1

a) Astronomical Telescope:-

It is a device used to observe a distinct images of heavily bodies.

It consists of a two convex lenses.

• Objective

having large focal length and large aperture.

• Eye piece

having small focal length and small aperture.

→ In normal adjustment final image is formed at infinity.

(as shown in figure)

Working:- The objective lens 'O' forms image A'B' of object which is at infinity. The image formed is real, diminished and inverted.

Now, the formed image acts as an object for eye piece or eye lens denoted by 'E' which forms final image at infinity. The final image which is formed is magnified and inverted.

Maginification of a telescope in normal adjustment :-

It is defined as ratio of angle subtended by the image at the eye piece and the angle subtended by the object at the objective when both object and image are at infinity.

i.e.

$\sf{M\:=\:\dfrac{\beta}{\alpha}\:=\:\dfrac{tan\:\beta}{tan\:\alpha}}$

$\sf{M\:=\: \dfrac{ \frac{A'B'}{O_2B'} }{\frac{A'B'}{O_1B'} }\:=\:\dfrac{O_1B'}{O_2B'}}$

$\sf{M\:=\:\dfrac{f_o}{-f_e}\:=\:\dfrac{-f_o}{f_e}}$

b)

From above

$\sf{M\:=\:\dfrac{f_o}{-f_e}\:=\:\dfrac{-f_o}{f_e}}$

We know that, P = 1/f

So,

$\sf{M\:=\:\dfrac{f_o}{-f_e}\:=\:\dfrac{-f_o}{f_e}\:=\:\dfrac{-P_e}{P_o}}$

According to question,

$\sf{P_e\:=\:10D}$ and $\sf{P_o\:=\:1D}$

Substitute the values in above formula

⇒ $\sf{M\:=\:\dfrac{-10D}{1D}}$

⇒ $\sf{M\:=\:-10}$

Therefore, magnifying power in normal adjustment is - 10.

Answer 2

❁❁ Refer To Image First .. ❁❁

An astronomical telescope is made up of two convex lenses :----

1) an objective lens.

ii) an eye piece .

The focal length fo of the objective lens of astronomical telescope is large as compared to the focal length fe of the eye piece.

working :---

→ A parallel beam of light from a star or a satellite falls on the objective lens of the telescope.

→ The objective lens forms a real, inverted and diminished image of the heavenly body.

→ This image now acts as an object for the eye piece, whose position is adjusted so that the image lies between the focus fe’ and the optical centre of the eye piece.

→ Now the eye piece forms a virtual, inverted and highly magnified image of object at infinity.

→ When the final image of an object is formed at infinity, the telescope is said to be in normal adjustment.

__________________________

Question :-- An astronomical telescope uses two lenses of power 10 D and 1 D the what is the magnifying power in normal adjustment ?

Given :--

→ f(e) = 1/10 = 0.1m = 10cm.

→ f(o) = 1/1 = 1m = 100cm.

Now, we know that, magnifying power(M) = -f(o)/f(e) ..

Putting values we get,

→ M = -(100/10)

→ M = -(10)

Hence, Magnifiying Power in normal adjustment is (-10) ...