Answer

Verified

414.6k+ views

**Hint:**Try to recall what normal adjustment is. In other words, when a telescope is set to normal adjustment the object distance becomes the sum of the focal lengths of the objective lens and the eye-piece. Substituting this in the lens formula should get you going.

**Formula used:**

Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} $ where $f$ is the focal length, $u$ is the object distance and $v$ is the image distance.

Magnification $M = \dfrac{Image\; height}{Object\;height}$ or $M = -\;\dfrac{v}{u}$.

Object distance $u = - (f_0 +f_e)$ where $f_0$ is the focal length of the objective lens and $f_e$ is the focal length of the eye-piece (applicable only here).

**Complete step-by-step answer:**

Let us deconstruct the question and understand what it means.

So, we have an astronomical telescope and we draw a straight line on the inside part of the objective lens. This line becomes the object for the eye-piece.

Now let us understand what normal adjustment is. When a telescope is adjusted so that the final image of a distant object is at infinity so that the eye is completely relaxed while viewing it is called normal adjustment. This is achieved when the distance between the objective and the eyepiece is equal to their focal lengths added together. The image formed by the objective lens will now lie between the principal focus and the optical centre of the eye lens instead of laying at the focus of the eye lens. This creates a virtual image at a finite distance from the eye lens.

From the diagram we can deduce that the object distance $u = - (f_0 +f_e)$. The negative sign is a convention for the object distance of all lenses and mirrors.

Substituting the object distance in the lens formula with respect to the eyepiece:

$\dfrac{1}{f} = \dfrac{1}{v} -\dfrac{1}{u} \Rightarrow \dfrac{1}{f_e} = \dfrac{1}{v} - \dfrac{1}{- (f_0 +f_e)} \Rightarrow \dfrac{1}{v} = \dfrac{1}{f_e} - \dfrac{1}{(f_0 +f_e)}$

$\Rightarrow \dfrac{1}{v} = \dfrac{f_0+f_e-f_e}{f_e(f_0+f_e)} \Rightarrow \dfrac{1}{v} = \dfrac{f_0}{f_e(f_0+f_e)} \Rightarrow v = \dfrac{ f_e(f_0+f_e)}{f_0} $

We know that magnification $M = \dfrac{Image\; height}{Object\;height}$ or $M = -\;\dfrac{v}{u}$. Equating the two and substituting line lengths:

$ -\dfrac{v}{u} = \dfrac{Image\; height}{Object\;height} \Rightarrow \dfrac{l}{L} = -\; \dfrac{\dfrac{ f_e(f_0+f_e)}{f_0}}{- (f_0 +f_e)} = \dfrac{f_e(f_0+f_e)}{f_0(f_0+f_e)} $

$\Rightarrow \dfrac{l}{L} = \dfrac{f_e}{f_0}$

But in normal adjustment, magnification is given as $M = \dfrac{f_0}{f_e} = \dfrac{L}{l}$

**So, the correct answer is “Option A”.**

**Note:**

Keep in mind that in accordance with the Cartesian sign convention, the object distances (u) are always negative since the object is placed to the left of the lens.

Also remember that for a telescope to be in normal adjustment the distance between the objective and the eyepiece should be equal to their focal lengths added together. Unless specified, you cannot assume the normal adjustment condition for all cases.

Recently Updated Pages

Differentiate between Shortterm and Longterm adapt class 1 biology CBSE

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Trending doubts

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

Write the 6 fundamental rights of India and explain in detail