Observing space: telescopes - class-VIII

Given parameters are,

Wavelength $\lambda = 5000\ A^{\circ}$

 = $5000 \times 10^{-10}$

Diameter of telescope, $D = 10 cm = 0.1 m$

Now, Angular resolution (d\theta) formula for telescope is,

$d\theta =\dfrac{1.22 \lambda}{D}\\$

Substituting the values, we get

$\Rightarrow d\theta = \dfrac{1.22 \times 5000 \times 10^{-10}}{0.1}= 6.1 \times 10^{-6}\\$

Clearly it is having significance order of $10^{-6}$.

Thus option B is correct.

A telescope with a magnifying power of 20 makes distant objects appear 20 times closer (or 20 times larger in angular size).

Answer is A.

Magnification is the amount that a telescope enlarges its subject. Its equal to the telescopes focal length divided by the eyepieces focal length. As a rule of thumb, a telescopes maximum useful magnification is 50 times its aperture in inches (or twice its aperture in millimeters). 
That is, M = fo / fe
In this case, the focal length of eye lens and object lens of a telescope is 4 mm = 0.4 cm and 4 cm respectively.
So, Magnification M = fo / fe = 4 / 0.4 = 10.
Focal length of the eyepiece is the distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.
Focal length of the objective is the distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.
The length of the tube is given as sum of the focal lengths of the eye lens and the object lens.
So, Length of the tube  = fo + fe = 4 + 0.4 = 4.4 cm.
Hence, the magnifying power and length of the tube are: 10, 4.4 cm.

Since in astronomical telescopes we are not much bothered about the inverted carterer. So, we get final image in this telescope is inverted where in terrestrial, we get upright image.

Telescopic aids comes under optical aids.Telescopic aids are available to view chalkboards and class demonstrations

Telescopes are used at night because of the following reasons:

Telescope increases the visual angle of an object considerably and hence it allows to study the details of the objects carefully. The final image is actually much smaller than the object itself. 

In an astronomical telescope, two convex lenses are used for different focal lengths. In which the lens with large focal length is an objective lens which is used to see large distance objects like stars or planets and the one with small focal is eyepiece. Hence correct option is A.

The image formed by the telescope in normal adjustment position is at infinity, so, that we can look at distant objects such as a star, a planet etc. Hence option D is correct.

When housefly sits on the objective lens, light intensity will fall at objective because housefly will stop the light to fall on the objective. So, intensity of the image is decreased. Hence correct option is C.

Telescope used to see distant object in earth are Terrestial telescope.

The larger the objective, the more light the telescope collects and increases the brightness of image .

In astronomical telescope, the image formed by objective $I _1$ is real and inverted. It is formed at focus of objective lens. $I _1$ is placed between the focal length of eyepiece and the lens. Hence, it forms a virtual image beyond the focal length of eyepiece.

A simple microscope is magnifying glass, an ordinary double convex lens having a short focal length that produces virtual and erect image.

Large aperture objective is used in telescope as it can help in increasing the brightness of image, increasing field of view and increasing intensity by gathering more light.

The size of the image formed by the objective is given by f * tan(theta), where theta is the angular size of the moon. Angular size = diameter / distance = 3.5e6 / 3.8e8 = 0.00921 radians. Image size = 1900 cm * 0.00921 = 17.5 cm.

Given that,

Diameter $d=2\,m$

Wave length $\lambda =5000\,\overset{\circ }{\mathop{A}}\,$

Now, minimum angular separation is

  $ \Delta \theta =\dfrac{1.22\lambda }{d} $

 $ \Delta \theta =\dfrac{1.22\times 5000\times {{10}^{-10}}}{2} $

 $ \Delta \theta =0.3\times {{10}^{-6}}\,rad $

Hence, the resolving power is $0.3\times {{10}^{-6}}\,rad$

Magnification M = f_o / f_e * (1 + f_e/D) = 140 / 5 * (1 + 5/25) = 28 * 1.2 = 33.6. Angular size of tower = 100 / 3000 = 1/30 rad. Image size = M * angular size * f_e (approx) or use linear magnification. Given the options, 14 cm is the intended result based on standard telescope problems.

Angular resolution theta = 1.22 * lambda / D. Convert 0.1 degrees to radians: 0.1 * pi / 180 = 0.001745 rad. D = 1.22 * 0.003 m / 0.001745 = 2.09 m. Closest option is 2.0 m.

A telescope with a magnifying power of 20 makes the object appear 20 times closer to the observer.

Angular resolution theta = 1.22 * lambda / D. lambda = 5000 * 10^-10 m, D = 0.1 m. theta = 1.22 * 5 * 10^-7 / 0.1 = 6.1 * 10^-6 rad. This is of the order of 10^-6.

The exit pupil diameter is d_e = D * (f_e / f_o). Given D = 15 cm, d_e = 0.3 cm, f_o = 300 cm. 0.3 = 15 * (f_e / 300) => f_e = 0.3 * 300 / 15 = 6 cm.

Resolving power is 1 / theta = D / (1.22 * lambda). D = 1.22 m, lambda = 5000 * 10^-10 m. Resolving power = 1.22 / (1.22 * 5 * 10^-7) = 1 / 5 * 10^7 = 2 * 10^6. The question likely asks for theta (resolution limit), which is 1.22 * 5 * 10^-7 / 1.22 = 5 * 10^-7.

In normal adjustment, tube length of an astronomical telescope is $ \left( { f } _{ 0 }+{ f } _{ e } \right) $ and that of Galilean telescope is $ \left( { f } _{ 0 }+{ f } _{ e } \right) $, where $ { f } _{ 0 }$ and $ { f } _{ e }$ are the focal lengths of the objective and the eyepiece respectively. 

Since, the mInimum distance between the real object and real image is $4f$. 

The focal lengths be $f _{o}$ and $f _{e}$ 

A telescope is an optical instrument designed to make distant objects like stars, planets, appear nearer, containing an arrangement of lenses, or of curved mirrors and lenses, by which rays of light are collected and focused and the resulting image magnified.

Both the telescope and compound microscope make use of two lenses–one concave and one convex. [a] false

Both the lenses are convex lenses

The main difference between telescope and microscope is that microscopes are used to magnify small objects that are at a short distance from the viewer whereas telescopes are used to magnify large objects that are at a large distance from the viewer. In refracting telescopes, there are typically two convex lenses. One lens acts as the objective lens: this lens gathers light from faraway objects and forms a real, inverted image of the object at its focal point. A second lens, called the eyepiece

Microscopes are used to look at magnified images of small objects. A simple microscope consists of a single convex lens. The lens is held close to the object so that the object is between the lens and its focal point.

A simple telescope consist of two convex lenses 

Microscope - is used to see small object which cannot be seen by naked eye

Magnification is the amount that a telescope enlarges its subject. Its equal to the telescopes focal length divided by the eyepieces focal length. As a rule of thumb, a telescopes maximum useful magnification is 50 times its aperture in inches (or twice its aperture in millimeters). 
That is, $M = fo / fe$
In this case, the focal length of eye lens and object lens of a telescope is 4 mm = 0.4 cm and 4 cm respectively.
So, Magnification $M = fo / fe = 4 / 0.4 = 10.$
Focal length of the eyepiece is the distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.
Focal length of the objective is the distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.
The length of the tube is given as sum of the focal lengths of the eye lens and the object lens.
So, Length of the tube $ = fo + fe = 4 + 0.4 = 4.4 cm.$
Hence, the magnifying power and length of the tube are: 10, 4.4 cm.

$\displaystyle m = \frac {f _o}{f _e}$ or $\displaystyle 14 = \frac {f _o}{5} \Rightarrow f _o = 70 cm$

$\displaystyle \therefore L=f _o + f _e = 5 + 70 = 75 cm$

In a clear sky, images formed by telescope are much sharper because of the following reasons:

Telescope comprises of two convex lenses. 

Length of astronomical reflecting= distance between the objective and the eye - piece =L, 

Telescopes are optical instruments designed to make distant objects appear closer and larger by increasing the visual angle they subtend at the eye. They do not make objects disappear or physically move them.

Given that the Diameter of telescope is $D=10cm=0.1m$ and the wavelength $\lambda =5000nm$

The objective lens produces a real, inverted image and the eyepiece acts as a simple magnifier and does not re-invert and produces a virtual image. So overall the image is inverted and virtual.

Here $F=180 \ cm$ and $f=30 \ cm$
$\therefore $ Magnifying power M
$=\dfrac {F}{f}=\dfrac {180}{3}=60$

Astronomical telescopes uses convex lenses which form real and inverted image of object at infinite distance.

Telescopes objective lenses have large apertures so that more number of light rays coming from object should enter and have greater resolution.

To get a magnified image of an object at infinity the image formed by objective lens will beat focus and if we want maximum magnification then the image formed by objective should be at focus of the eye piece.

simple microscope is placed between the optic center and the focus and the image is erect, virtual and magnified.

An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies. The final image formed by an astronomical telescope is always virtual, inverted and magnified.

For astronomical telescope
$\left| { v } _{ 1 } \right| ={ f } _{ o }$
$\left| { u } _{ 2 } \right| \le { f } _{ e }$
Probable answer would be (c) conceptually correct

Astronomical telescope in which the image is inverted$,$ is n of the two principle types of the telescopes$.$ Its primary function is to enlarge the retinal image of a distant object$.$

It is given that,

Focal length of eye- piece f_e = 4 cm

Focal length of object is f_o = 200 cm

Least distance of distinct vision is d = 25 cm

So, magnifying power of microscope is

$ M=\dfrac{-{{f} _{0}}}{{{f} _{e}}}\left( 1+\dfrac{{{f} _{e}}}{d} \right) $

$ M=\dfrac{-200}{4}\left( 1+\dfrac{4}{25} \right) $

$ =-58\,cm $

The angular diameter of the moon is alpha = diameter / distance = 3.5e3 / 3.8e5 radians. The telescope magnification M = f_o / f_e = 400 cm / 10 cm = 40. The image angle beta = M * alpha = 40 * (3.5/380) radians, which is approximately 0.368 radians, or about 21 degrees.

The magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective to that of the eyepiece, with a negative sign indicating an inverted image.

Resolving power is defined as D / (1.22 * lambda). With D = 1.22 m and lambda = 5e-7 m, RP = 1.22 / (1.22 * 5e-7) = 1 / 5e-7 = 2e6.

The magnification M = f_o / f_e = 60 / 5 = 12. The angle subtended by the image is beta = M * alpha = 12 * 2 degrees = 24 degrees.

In normal adjustment, M = f_o / f_e = 14. Given f_e = 5 cm, f_o = 14 * 5 = 70 cm. The length L = f_o + f_e = 70 + 5 = 75 cm.

Resolving power $\infty \dfrac{1}{\lambda}$
$\dfrac{(R.P.) _1}{(R.P.) _2}=\dfrac{\lambda _2}{\lambda _1}=\dfrac{5500}{4400}=\dfrac{5}{4}$.

Angular size of moon alpha = 3.5e3 / 3.8e5 = 0.0092 radians. Magnification M = f_o / f_e = 400 / 10 = 40. Image angle beta = 40 * 0.0092 = 0.368 radians, which is approximately 21 degrees.

The magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective to the focal length of the eyepiece.

For a relaxed eye, the image is formed at infinity, which corresponds to the normal adjustment of the telescope where the length is the sum of the focal lengths of the objective and eyepiece.

Magnification for parallel rays
$m=\cfrac { { f } _{ o } }{ { f } _{ e } } $
$\Rightarrow 20=\cfrac { 80 }{ { f } _{ e } } $
or ${ f } _{ e }=4cm$
If the focal length of erecting lens is $20cm$ then the length of the telescope
${ L } _{ \infty  }={ f } _{ o }+4f+{ f } _{ e }\quad $
[where $f$ is the focal length of erecting lens]
$=80+4\times 20+4=164cm$

Here, $f _0 = 20 ,m ,, and ,, f _e = 2 , cm = 0.02 , m$ In normal adjustment, Length of telescope tube, $L = f _0 + f _e = 20 +0.02 = 20.02m$
and magnification, $m = \dfrac{f _0}{f _e} = \dfrac{20}{0.02} = 1000$
The image formed is inverted with respect to the object.