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Question: A monochromatic light passes through a glass slab \(\left( {\mu \, = \,1.5} \right)\) of thickness \...

A monochromatic light passes through a glass slab (μ = 1.5)\left( {\mu \, = \,1.5} \right) of thickness 9 cm9\,cm in time t1{t_1}. If it takes a time t2{t_2} to travel the same distance through water (μ = 43)\left( {\mu \, = \,\dfrac{4}{3}} \right). The value of (t1 − t2)\left( {{t_1}\, - \,{t_2}} \right) is:
A. 5 × 10−11 sec⁡5\, \times \,{10^{ - 11}}\,\sec
B. 5 × 10−8 sec⁡5\, \times \,{10^{ - 8}}\,\sec
C. 2.5 × 10−10 sec⁡2.5\, \times \,{10^{ - 10}}\,\sec
D. 5 × 10−10 sec⁡5\, \times \,{10^{ - 10}}\,\sec

Explanation

Solution

We can easily reach to the solution to this question by first considering what the refractive index actually means, then finding the velocity of light in each medium and then using it to find the difference between the times taken by light in passing the same difference through both mediums.

Complete step-by-step solution:
We will try to solve the question exactly as mentioned in the hint section of the solution, without committing any mistakes, this question can be easily solved as follows:
Firstly, we need to know about the refractive index and what it means. We know that we can define refractive index mathematically as:
μ = cv\mu \, = \,\dfrac{c}{v}
where cc is already known to us as the speed of light in vacuum and,
vv is the speed of light in a medium
We already know that:
t = svt\, = \,\dfrac{s}{v}
Where, tt is the time taken to travel a distance ss with the velocity of vv .
It is given that refractive index of glass slab is,
μ1 = 1.5 = 32{\mu _1}\, = \,1.5\, = \,\dfrac{3}{2}
Using the equation of refractive index:
v1 = cμ1{v_1}\, = \,\dfrac{c}{{{\mu _1}}}
Substituting the values of cc and μ1{\mu _1} that are known to us, we get:
v1 = 2 × 108 m/s{v_1}\, = \,2\, \times \,{10^8}\,m/s
Distance that needs to be travelled is, d = 9 cm = 9×10−2 md\, = \,9\,cm\, = \,9 \times {10^{ - 2}}\,m
t1 = dv1 = 9×10−22×108 = 4.5 × 10−10 s{t_1}\, = \,\dfrac{d}{{{v_1}}}\, = \,\dfrac{{9 \times {{10}^{ - 2}}}}{{2 \times {{10}^8}}}\, = \,4.5\, \times \,{10^{ - 10}}\,s
⇒ t1 = 4.5 × 10−10 s\Rightarrow \,{t_1}\, = \,4.5\, \times \,{10^{ - 10}}\,s
Now, we will discuss about the situation in water:
It is given that refractive index of water is,
μ2 = 43{\mu _2}\, = \,\dfrac{4}{3}
Using the equation of refractive index:
v2 = cμ2{v_2}\, = \,\dfrac{c}{{{\mu _2}}}
Substituting the values of cc and μ1{\mu _1} that are known to us, we get:
v2 = 2.25 × 108 m/s{v_2}\, = \,2.25\, \times \,{10^8}\,m/s
Distance that needs to be travelled is already known to us as d = 9 cm = 9×10−2 md\, = \,9\,cm\, = \,9 \times {10^{ - 2}}\,m
t2 = dv2 = 9×10−22.25×108 = 4 × 10−10 s{t_2}\, = \,\dfrac{d}{{{v_2}}}\, = \,\dfrac{{9 \times {{10}^{ - 2}}}}{{2.25 \times {{10}^8}}}\, = \,4\, \times \,{10^{ - 10}}\,s
⇒ t2 = 4 × 10−10 s\Rightarrow \,{t_2}\, = \,4\, \times \,{10^{ - 10}}\,s
The question asked us the difference of the time taken by the light to travel in both mediums, so:
t1 − t2 = (4.5 − 4) ×10−10 s{t_1}\, - \,{t_2}\, = \,\left( {4.5\, - \,4} \right)\, \times {10^{ - 10}}\,s
⇒ (t1 − t2) = 5 ×10−11 s\Rightarrow \,\left( {{t_1}\, - \,{t_2}} \right)\, = \,5\, \times {10^{ - 11}}\,s
Hence, the correct answer is the option (A).

Note:- Many students make mistakes in the part where they have to manipulate what refractive index is and use it to find the time taken to cover the given distance. Even if they do it the right way, calculation mistakes can be made because refractive indices may sometimes be confusing.