The distance of a galaxy is $56\times {{10}^{25}}m$ . Assume the speed of light to be $3\times {{10}^{8}}m{{s}^{-1}}$ . Find the time taken by light to travel this distance.
a)$0.87\times {{10}^{18}}s$
b)$1.87\times {{10}^{18}}s$
c)$18.7\times {{10}^{18}}s$
d)$1.87\times {{10}^{17}}s$
Answer
595.2k+ views
Hint: In the question the distance from a particular galaxy is given as $56\times {{10}^{25}}m$ . The light travels with a constant speed in a medium and for every medium it is a constant. Hence using the relation between the speed and the distance covered by the object, the time taken can be determined accordingly.
Formula used:
$s=\dfrac{d}{t}$
Complete answer:
Let us say a particle travels with a constant speed ‘s’ in a particular medium. If the particle covers a distance ‘d’ in time ‘t’, than the relation between the above parameters of motion is given by the equation,
$s=\dfrac{d}{t}$
Light exhibits a dual nature i.e. it can be treated to be a wave as well at the same time it can be treated as a particle. For the above problems let us consider the light to be consisting of particles called photons. Let us say a photon leaves the galaxy and travels with a speed of $3\times {{10}^{8}}m{{s}^{-1}}$. The distance from the galaxy is given as $56\times {{10}^{25}}m$ . hence using the above relation the time taken by the photon to reach Earth or a reference point is equal to,
$\begin{align}
& s=\dfrac{d}{t} \\
& \Rightarrow 3\times {{10}^{8}}m{{s}^{-1}}=\dfrac{56\times {{10}^{25}}m}{t} \\
& \Rightarrow t=\dfrac{56\times {{10}^{25}}m}{3\times {{10}^{8}}m{{s}^{-1}}} \\
& \therefore t=1.87\times {{10}^{18}}s \\
\end{align}$
Therefore the correct answer of the above question is option b.
Note:
Between the galaxy and the Earth there is nothing but outer space. Outer space is perfectly not vacuum, but it can always be treated equivalent to a vacuum. Light travels with a maximum speed in vacuum as the refractive index of vacuum is one. As the refractive index of a medium increases i.e. mediums that are optically denser, light travels with least speed in such mediums.
Formula used:
$s=\dfrac{d}{t}$
Complete answer:
Let us say a particle travels with a constant speed ‘s’ in a particular medium. If the particle covers a distance ‘d’ in time ‘t’, than the relation between the above parameters of motion is given by the equation,
$s=\dfrac{d}{t}$
Light exhibits a dual nature i.e. it can be treated to be a wave as well at the same time it can be treated as a particle. For the above problems let us consider the light to be consisting of particles called photons. Let us say a photon leaves the galaxy and travels with a speed of $3\times {{10}^{8}}m{{s}^{-1}}$. The distance from the galaxy is given as $56\times {{10}^{25}}m$ . hence using the above relation the time taken by the photon to reach Earth or a reference point is equal to,
$\begin{align}
& s=\dfrac{d}{t} \\
& \Rightarrow 3\times {{10}^{8}}m{{s}^{-1}}=\dfrac{56\times {{10}^{25}}m}{t} \\
& \Rightarrow t=\dfrac{56\times {{10}^{25}}m}{3\times {{10}^{8}}m{{s}^{-1}}} \\
& \therefore t=1.87\times {{10}^{18}}s \\
\end{align}$
Therefore the correct answer of the above question is option b.
Note:
Between the galaxy and the Earth there is nothing but outer space. Outer space is perfectly not vacuum, but it can always be treated equivalent to a vacuum. Light travels with a maximum speed in vacuum as the refractive index of vacuum is one. As the refractive index of a medium increases i.e. mediums that are optically denser, light travels with least speed in such mediums.
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