Tag: observing space: telescopes

Quizzes (9) Questions (139)

Questions Related to observing space: telescopes

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

The wavelengths for the light of red and blue colours are roughly $7.8\times 10^{-7}m$ and $4.8\times 10^{-7}m$ respectively. Which colour has the greater speed in glass?

  1. Red light

  2. Violet light

  3. Yellow light

  4. All colors have same speed
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When travelling through a medium, the frequency of light doesn't change. Hence, red light having the maximum wavelength will travel fastest through glass as speed of light is given by $v=f\lambda$ where; $f$ is the frequency and $\lambda$ is the wavelength of light travelling at speed $v$.

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

What is the speed of Light $c$?

  1. $c=3\times 10^{-8} m s^{-1}$
  2. $c=3\times 10^8 km s^{-1}$
  3. $c=3\times 10^8 mm s^{-1}$
  4. $c=3\times 10^8 m s^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its value is exactly $299792458$ meters per second.
That is, $c=3\times { 10 }^{ 8 }m/s$.

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

In which form does the Light propagates?

  1. Electric waves

  2. Magnetic waves

  3. Electromagnetic waves

  4. Sound wave
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electromagnetic waves are waves which can travel through the vacuum of outer space.light waves are examples of electromagnetic waves.
so,Light propagates in the form of electromagnetic waves.

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

As speed decreases, if we change medium of electromagnetic waves from air to water, frequency

  1. also decreases.

  2. also increases.

  3. remains same.

  4. may increase or decrease.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As speed decreases, if change medium of electromagnetic wave from air to water, frequency remains same.

As we go from one medium to another the frequency stays same, but the wavelength changes. Change in speed is proportional to change in wavelength.

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

A plane electromagnetic wave of angular frequency ${ mm }^{ 2 }$ propagates in a poorly conducting medium of conductivity $\sigma $ and relative permittivity $\epsilon $ Find the ratio of conduction current density and displacement current density in the medium.

  1. ${ \epsilon \epsilon } _{ 0 }\omega /\sigma $
  2. $\sigma /{ \epsilon \epsilon } _{ \sigma }\omega $
  3. $\omega /{ \sigma \epsilon } _{ 0 }\omega $
  4. $\omega \sigma /{ \epsilon } _{ 0 }\epsilon $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

Which among the four options is not correct ion given situation ? A charged particle oscillates about its mean equilibrium position with a frequency of $10^9\, Hz.$ The electromagnetic waves produced 

  1. Will have frequency of $10^9\, Hz.$
  2. Will have frequency of $2\times 10^9\, Hz.$
  3. Will have wavelength of $0.3\,m.$

  4. Fall in the region of radio waves
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An oscillating charge produces electromagnetic waves of the same frequency as the oscillation. Wavelength lambda = c / f. With f = 10^9 Hz and c = 3 * 10^8 m/s, lambda = 0.3 m. Thus, option C is correct, while A and B are incorrect.

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

An electromagnetic wave with frequency $\omega$ and wavelength $\lambda$ travels in the $+y$ direction. Its magnetic field is along $-x$ axis. The vector equation for the associated electric field (of amplitude E_0)is :

  1. $ \vec { E } = E _0 \cos \left(\omega t -\dfrac{2\pi}{\lambda}y\right)\hat{x}$
  2. $\vec{ E } = -E _0 \cos \left(\omega t +\dfrac{2\pi}{\lambda}y\right)\hat{x}$
  3. $-\vec{ E } = E _0 \cos \left(\omega t +\dfrac{2\pi}{\lambda}y\right)\hat{z}$
  4. $\vec { E } = E _0 \cos \left(\omega t -\dfrac{2\pi}{\lambda}y\right)\hat{z}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

Electro magnetic waves are transverse in nature as is evident by

  1. polarization

  2. interference

  3. reflection

  4. deffraction
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electromagnetic waves are transverse in nature as is evident by polarization interference and diffraction explain the wave nature of light or $EM$ waves polarization is the phenomenon by which we restrict the vibration of wave in a particular direction perpendicular to direction of $W$ are propagation. So, it explains the transverse nature of $EM$ waves.

Answer :- polarization.

Multiple choice the nature of electromagnetic waves space exploration and forms of light observing space: telescopes electromagnetic waves physics

The maximum kinetic energy of electron if wavelength of incident electromagnetic wave is $260 \,nm$ and cut-off wavelength is $380 \,nm$ given $hc = 1237 \,nm-eV$ is

  1. $1.5 \,eV$
  2. $6.4 \,eV$
  3. $10 \,eV$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know 
$KE _{max} = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda _0} $
$= \dfrac{1237 \, nm. eV}{260 \, nm} - \dfrac{1237 \, nm . eV}{380 \, nm}$
$= 1.5 eV$