
A glass slab of thickness t and refractive index $\mu $. Calculate the time taken by light to travel this thickness.
$
(a){\text{ t}}\mu {\text{c}} \\
{\text{(b) }}\dfrac{{tc}}{\mu } \\
(c){\text{ }}\dfrac{t}{{\mu c}} \\
(d){\text{ }}\dfrac{{\mu t}}{c} \\
$
Answer
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Hint – In this question use the concept that the refractive index of glass slab will be equal to the ratio of speed of light in air to that of speed of light in that medium that is $\mu = \dfrac{c}{v}$. The basic relationship between the distance, speed and time that is $d = v \times t$ will help approaching the problem.
Formula used - $\mu = \dfrac{c}{v}$, $d = v \times t$
Complete step-by-step answer:
Given data:
Thickness of the glass slab = t
Refractive index of the glass slab = $\mu $
Let the time taken by the light to travel in the thickness of the glass slab = T.
Now as we know that the velocity of light in the medium is equal to the ratio of velocity of light in the air to the refractive index of the medium.
Let velocity of light in the medium (i.e. through the thickness of the glass slab) be v m/s.
Velocity of light in the air be C = $3 \times {10^8}$m/s.
So the velocity of light in the thickness of the glass slab = $\dfrac{C}{\mu }$.
Now as we know the relation between the velocities (v), distance (d) and time (t) taken is
$v = \dfrac{d}{t}$.......... (1) (i.e. velocity is the ratio of distance travelled divided by the time taken to cover this distance)
Here the distance is the thickness of the glass slab.
Therefore, d = t
Velocity is the velocity of light in the medium.
Therefore, $v = \dfrac{C}{\mu }$
And time (t) is the time taken by the light to travel in the thickness of the glass slab.
Therefore, t = T.
Now substitute these values in equation (1) we have,
$ \Rightarrow \dfrac{C}{\mu } = \dfrac{t}{T}$
Now simplify it and calculate the value of time taken (T)
$ \Rightarrow T = \dfrac{{\mu t}}{C}$
So this is the required time taken by the light to travel in the thickness of the glass slab.
Hence option (D) is the correct answer.
Note – The trick point here was that the distance travelled by the light in the glass slab will be equal to the thickness of the glass slab. The value of light in air is always constant as is denoted by c that is $c = 3 \times {10^8} m/s$. It is advised to remember this value as it is very helpful in problems of this kind.
Formula used - $\mu = \dfrac{c}{v}$, $d = v \times t$
Complete step-by-step answer:
Given data:
Thickness of the glass slab = t
Refractive index of the glass slab = $\mu $
Let the time taken by the light to travel in the thickness of the glass slab = T.
Now as we know that the velocity of light in the medium is equal to the ratio of velocity of light in the air to the refractive index of the medium.
Let velocity of light in the medium (i.e. through the thickness of the glass slab) be v m/s.
Velocity of light in the air be C = $3 \times {10^8}$m/s.
So the velocity of light in the thickness of the glass slab = $\dfrac{C}{\mu }$.
Now as we know the relation between the velocities (v), distance (d) and time (t) taken is
$v = \dfrac{d}{t}$.......... (1) (i.e. velocity is the ratio of distance travelled divided by the time taken to cover this distance)
Here the distance is the thickness of the glass slab.
Therefore, d = t
Velocity is the velocity of light in the medium.
Therefore, $v = \dfrac{C}{\mu }$
And time (t) is the time taken by the light to travel in the thickness of the glass slab.
Therefore, t = T.
Now substitute these values in equation (1) we have,
$ \Rightarrow \dfrac{C}{\mu } = \dfrac{t}{T}$
Now simplify it and calculate the value of time taken (T)
$ \Rightarrow T = \dfrac{{\mu t}}{C}$
So this is the required time taken by the light to travel in the thickness of the glass slab.
Hence option (D) is the correct answer.
Note – The trick point here was that the distance travelled by the light in the glass slab will be equal to the thickness of the glass slab. The value of light in air is always constant as is denoted by c that is $c = 3 \times {10^8} m/s$. It is advised to remember this value as it is very helpful in problems of this kind.
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