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Q2.5:
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?

Solution :

Step 1: Establishing the System of Units

In the standard International System of Units (SI), the speed of light in vacuum is defined as $c \approx 3.00 \times 10^8 \, \text{m/s}$. According to the problem statement, we define a new system where the speed of light, $v_{new}$, is assigned a numerical value of unity.

$v_{new} = 1 \, \text{new unit of length/unit of time} = 1$

[Per the definition of velocity: $v = \frac{d}{t}$]

Step 2: Converting Time to Standard Units

The time interval ($t$) taken for light to travel from the Sun to the Earth is given as 8 minutes and 20 seconds. We must express this entirely in seconds to maintain chronological consistency.

$t = (8 \, \text{min} \times 60 \, \text{s/min}) + 20 \, \text{s}$

$t = 480 \, \text{s} + 20 \, \text{s} = 500 \, \text{s}$

Step 3: Calculating Distance in New Units

Using the kinematic equation for constant speed, $d = v \times t$. Given that the speed of light is defined as unity ($1$) in our new system, the distance $d$ is numerically equivalent to the time $t$.

$d = v_{new} \times t$

$d = 1 \times 500 = 500 \, \text{new units of length}$

Visual Representation of Light Propagation

Sun Earth d = 500 light-units

Step 4: Theoretical Justification

The distance covered by light in a specific duration, when speed is normalized to unity, effectively transforms the time-interval measurement into a spatial measurement. This is analogous to the astronomical definition of a light-second, where the unit of distance is the distance light travels in one second. Thus, $500$ seconds of travel time results in a distance of $500$ light-units. [By the principle of dimensional consistency in defining arbitrary units].

Final Solution: The distance between the Sun and the Earth in the new unit of length is 500 units.


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