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# Astronomical Telescope Assignment Help

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Optical Physics - Astronomical Telescope

**Astronomical Telescope**

An astronomical telescope is an optical instrument which is used for observing distinct images of heavenly bodies like stars, planets etc.

It consists of two lenses (or lens systems), the objective lens O, which is of large focal length and large aperture and the eye piece E, which has a small focal length and small aperture. The two lenses are mounted co-axially at the free ends of the two tubes. The distance between these lenses can be adjusted using a rack and pinion arrangement.

In normal adjustment of telescope, the final image is formed at infinity.

The course of rays in normal adjustment of telescope, a parallel beam of light from an astronomical object (at infinity) is made to fall on the objective lens of the telescope. It forms a real, inverted and diminished image A’B’ lies just at the focus of the eye piece. A’B’ lies just at the focus of the eye piece.

Therefore, a final highly magnified image is formed at infinity. The final image is erect w.r.t., A’B’ and is inverted w.r.t. the object.

However, in astronomical telescope, final image being inverted w.r.t. the objects are usually spherical.

Magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye, by the object directly, when the final image and the object both lie at infinite distance from the eye.

As the object lies at very huge distance, therefore, angle subtended by the object at **C**_{2} (where eye is held) is almost the same as the angle subtended by the object at **C**_{1} (because **C**_{1} is close to **C**_{2}). Let it be α, i.e. **∠A’C**_{1}B = α . Rays coming from the final image at infinity make **∠A’C**_{1}B’ = α on the eye. Therefore, by definition,

Magnifying power, **m = β/ α ** ** (1)**

As angles α and **β** are small, therefore, α **≈ tan ** α and **β ≈ tan β**.

From (1), **m = tan β/tan α ** ** (2)**

In **Δ A’B’C**_{2}, tan β = A’B’/C_{2}B’

As the object is very far off, angle subtended by the object on the eye is almost the same as the angle subtended by it at **C**_{1}. Let it be α ,

Therefore, **∠ A’C**_{1}B’ = α

Further, let **∠A”C**_{2}B” = β,

Where **C2B” = β**,

Where **C**_{2}B” = d

Therefore, by definition, magnifying power,

**M = β/ α (3)**

As the angles α and **β** are small, therefore, **β ≈ tan β and tan α ≈ tan ** α

From (3), **m = tan β/tan α (4)**

In **Δ A’B’C**_{2}, tan β = A’B’/C_{2}B’

In **Δ A’B’C**_{1}, tan α = A’B’/C_{1}B’

Putting in (4), we get,

**m = (A’B’/C**_{2}B’) × C_{1}B’/A’B’

**m = C**_{1}B'/C_{2}B’ = ƒ_{0}/-u_{e} (5)

where **C**_{1}B’ = ƒ_{0} = focal length of objective lens **C**_{2}B’ = -u_{e}, distance of A’B’, acting as the object for eye lens.

Now, for eye lens, **1/v – 1/u = 1/ƒ**

Taking **v**_{e} = -d, u = -u_{e}

And **ƒ = + ƒ**_{e}, we get

**1/-d – (1/-u**_{e}) = 1/ƒ_{e}

**1/u**_{e} = 1/ƒ_{e} + 1/d = 1/ƒ_{e} [1 + (ƒ_{e}/d)]

Putting in (5), we get,

**M = ƒ**_{0}/ƒ_{e} [1 + (ƒ_{e}/d)]

**Discussion:**

**(i) **As magnifying power is negative, the final image in an astronomical telescope is inverted i.e. upside down and left turned right.

**(ii) **As intermediate image is between the two lenses, cross wire (or a measuring device) can be used.

**(iii) **In normal setting of telescope, final image is at infinity. Magnifying power is minimum.

When final image is at least distance of distinct vision, magnifying power is minimum. Thus

**(M.P.)**_{min} = -[ ƒ0/ƒe];

**(M.P.)**_{max} = ƒ_{0}/ƒ_{e} [(1 + (ƒ_{e}/d)]

**(iv) **To have large magnifying power, **ƒ**_{0} must be as possible and **ƒ**_{e} must be as small as possible.

**(v)** In a telescope, aperture of objective lens is made large to increase magnifying power and resolving power of the telescope.

**(vi) **The largest lens objective in use has a diameter of **102 cm**. this telescope at the Yerkes Observatory in Wisconsin (U.S.A).

**Note: **in a refracting astronomical telescope, distance between objective lens and eye is adjusted by moving the rack and pinion arrangement. This focuses the object situated at infinity.

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**Astronomical Telescope**

It consists of two lenses (or lens systems), the objective lens O, which is of large focal length and large aperture and the eye piece E, which has a small focal length and small aperture. The two lenses are mounted co-axially at the free ends of the two tubes. The distance between these lenses can be adjusted using a rack and pinion arrangement.

In normal adjustment of telescope, the final image is formed at infinity.

The course of rays in normal adjustment of telescope, a parallel beam of light from an astronomical object (at infinity) is made to fall on the objective lens of the telescope. It forms a real, inverted and diminished image A’B’ lies just at the focus of the eye piece. A’B’ lies just at the focus of the eye piece.

Therefore, a final highly magnified image is formed at infinity. The final image is erect w.r.t., A’B’ and is inverted w.r.t. the object.

However, in astronomical telescope, final image being inverted w.r.t. the objects are usually spherical.

Magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye, by the object directly, when the final image and the object both lie at infinite distance from the eye.

As the object lies at very huge distance, therefore, angle subtended by the object at

**C**

_{2}(where eye is held) is almost the same as the angle subtended by the object at

**C**

_{1}(because

**C**

_{1}is close to

**C**

_{2}). Let it be α, i.e.

**∠A’C**α . Rays coming from the final image at infinity make

_{1}B =**∠A’C**α on the eye. Therefore, by definition,

_{1}B’ =Magnifying power,

**m = β/ α**

**(1)**

As angles α and

**β**are small, therefore, α

**≈ tan**α and

**β ≈ tan β**.

From (1),

**m = tan β/tan α**

**(2)**

In

**Δ A’B’C**

_{2}, tan β = A’B’/C_{2}B’As the object is very far off, angle subtended by the object on the eye is almost the same as the angle subtended by it at

**C**

_{1}. Let it be α ,

Therefore,

**∠ A’C**α

_{1}B’ =Further, let

**∠A”C**

_{2}B” = β,Where

**C2B” = β**,

Where

**C**

_{2}B” = dTherefore, by definition, magnifying power,

**M = β/ α (3)**

As the angles α and

**β**are small, therefore,

**β ≈ tan β and tan α ≈ tan**α

From (3),

**m = tan β/tan α (4)**

In

**Δ A’B’C**

_{2}, tan β = A’B’/C_{2}B’In

**Δ A’B’C**

_{1}, tan α = A’B’/C_{1}B’Putting in (4), we get,

**m = (A’B’/C**

_{2}B’) × C_{1}B’/A’B’**m = C**

_{1}B'/C_{2}B’ = ƒ_{0}/-u_{e}(5)where

**C**focal length of objective lens

_{1}B’ = ƒ_{0}=**C**

_{2}B’ = -u_{e}, distance of A’B’, acting as the object for eye lens.

Now, for eye lens,

**1/v – 1/u = 1/ƒ**

Taking

**v**

_{e}= -d, u = -u_{e}

And

**ƒ = + ƒ**

_{e}, we get

**1/-d – (1/-u**

_{e}) = 1/ƒ_{e}

**1/u**

_{e}= 1/ƒ_{e}+ 1/d = 1/ƒ_{e}[1 + (ƒ_{e}/d)]Putting in (5), we get,

**M = ƒ**

_{0}/ƒ_{e}[1 + (ƒ_{e}/d)]**Discussion:**

**(i)**As magnifying power is negative, the final image in an astronomical telescope is inverted i.e. upside down and left turned right.

**(ii)**As intermediate image is between the two lenses, cross wire (or a measuring device) can be used.

**(iii)**In normal setting of telescope, final image is at infinity. Magnifying power is minimum.

When final image is at least distance of distinct vision, magnifying power is minimum. Thus

**(M.P.)**;

_{min}= -[ ƒ0/ƒe]**(M.P.)**

_{max}= ƒ_{0}/ƒ_{e}[(1 + (ƒ_{e}/d)]**(iv)**To have large magnifying power,

**ƒ**

_{0}must be as possible and

**ƒ**

_{e}must be as small as possible.

**(v)**In a telescope, aperture of objective lens is made large to increase magnifying power and resolving power of the telescope.

**(vi)**The largest lens objective in use has a diameter of

**102 cm**. this telescope at the Yerkes Observatory in Wisconsin (U.S.A).

**Note:**in a refracting astronomical telescope, distance between objective lens and eye is adjusted by moving the rack and pinion arrangement. This focuses the object situated at infinity.

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