Tag: observing space: telescopes

Questions Related to observing space: telescopes

The electric field of a plane electromagnetic wave is given by
$\vec{E} = E _0 \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \cos (kz + \omega t)$
At $t = 0$, a positively charged particle is at the point $(x, y , z) = \left(0, 0 , \dfrac{\pi}{k} \right)$. If its instantaneous velocity at $(t = 0)$ is $v _0 \hat{k}$, the force acting on it due to the wave is :

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. parallel to $\hat{k}$

  2. parallel to $\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

  3. antiparallel to $\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

  4. zero

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Correct Option: B
Explanation:

$\vec{E} = E _0 \left(\dfrac{i + j}{\sqrt{2}}\right) \cos (kz + wt)$


$\therefore$ unit vector along electric field, $\vec{E} = \left(\dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \right)$


Direction of electromagnetic wave is in (-z) direction 

$\therefore \hat{C} = -\hat{k}$  wave direction.

Let $\hat{B} $ is the unit vector along the direction of magnetic field.

$\hat{B} = \hat{C} \times \hat{E} = -\hat{k} \times \left(\dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \right) = - \left(\dfrac{\hat{k} \times i + \hat{k} \times j}{\sqrt{2}} \right)$

$\hat{B} = -\left(\dfrac{\hat{j} + (-i)}{\sqrt{2}} \right)  = \left(\dfrac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$

$\vec{F _e} =$ electric force on the charge particle 

$\vec{F _e} = $ unit vector of electric force $= \dfrac{q \hat{E}}{|q\hat{E}|} = \hat{E}$

$\vec{F} _e = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

$\vec{F} _b = $ magnetic force $= q \vec{V} \times \vec{B} = q \left(V _0 \hat{k} \times \dfrac{i - j}{\sqrt{2}}\right)$

$\vec{F} _b = q V _0 \left[\dfrac{\hat{k} \times \hat{i} - \hat{k} \times \hat{j}}{\sqrt{2}} \right] = q V _0 \left[\dfrac{\hat{j} - (-\hat{i})}{\sqrt{2}}\right]$

$\vec{F} _b = q V _0 \left(\dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \right)$

$\vec{F} _{Net} = \hat{F} _e + \hat{F} _b = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} + \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} (\hat{i} + \hat{j})$

$\therefore \hat{F} _{Net} = $ unit vector $= \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

Option (B) is correct.

The Schrodinger equation for a free electron of mass m and energy E written in terms of the wave function $\psi $ is $\frac{d^2\psi}{dx^2}+\frac{8 \pi ^2mE}{h^2}\psi =0$. The dimensions of the coefficient $\psi$ of in the second term must be

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $[M^1L^1]$

  2. $[L^2]$

  3. $[L^{-2}]$

  4. $[M^1L^{-1}T^1]$

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Correct Option: C
Explanation:

By dimensional analysis the dimensions of each term in an equation must be the same. In the first term the second derivative with respect to distance x indicates the dimensions of the coefficient $\psi$ of to be $[L^{-2}]$ and hence the answer.

The angular resolution of a 10cm diameter telescope at a wavelength of $5000 A^0$ is of the order of -

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. $10^{4} rad$

  2. $10^{-6} rad$

  3. $10^{6} rad$

  4. $10^{2} rad$

Show answer
Correct Option: B
Explanation:

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.

An observer looks at a distant tree of height $10$ m with a telescope magnifying power of $20$. To the observer, the top appears

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. 10 times taller

  2. 10 times nearer

  3. 20 times taller

  4. 20 times nearer

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Correct Option: D

The focal length of eye lens and object lens of a telescope is 4 mm and 4 cm respectively. If final image of an far object is at $\infty $. Then the magnifying power and length of the tube are :

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. 10, 4.4 cm

  2. 4, 44 cm

  3. 44,10 cm

  4. 10, 44 cm

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Correct Option: A
Explanation:

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.

Which of the following is correct about astronomical telescope?

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. It consists of two diverging lenses

  2. Its objective is a concave lens

  3. Its eyepiece is a convex lens with greater focal length than the objective

  4. The final image in this telescope is inverted

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Correct Option: D
Explanation:

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 are used:

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. to view planets and other astronomical activities

  2. to view chalkboards and class demonstrations

  3. for performing computational tasks

  4. helps visuaaly challenged people in reading and writing

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Correct Option: A
Explanation:

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

Why is it advised to used telescope at night?

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. During day, sun rays are very strong and it is not possible to see distant objects.

  2. Night sky is clearer than day sky.

  3. Both A and B

  4. Neither A nor B

Show answer
Correct Option: C
Explanation:

Telescopes are used at night because of the following reasons:

1. Sun rays in day are very strong and other distant objects are not visible since the intensity of light from distant sources is much weaker.
2. Night sky is clearer than day sky primarily due to lesser pollution. This helps in formation of higher quality images at night.

Which of the following is true for a telescope?

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. Final image is real and inverted.

  2. It increases the size of the object.

  3. It increases the angle subtended by image at the eye of the observer.

  4. Smaller telescopes are more powerful

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Correct Option: C
Explanation:

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 the focal lengths of objective and eyepiece should respectively be :

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. large and small

  2. small and large

  3. equal

  4. too small are too large

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Correct Option: A
Explanation:

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.