We are given a monochromatic light passing through a glass slab and water, and we need to calculate the difference in time taken by light to travel through both mediums over the same distance.
Given:
• Refractive index of glass (μglass\mu_{\text{glass}}) = 1.5
• Refractive index of water (μwater\mu_{\text{water}}) = 43\frac{4}{3}
• Thickness of the slab and water = 9 cm = 0.09 m
• We need to calculate the difference in time taken by light to travel through the two mediums: Δt=t1−t2\Delta t = t_1 - t_2
Formula for time taken:
The time taken by light to travel a distance dd in a medium is given by:
t=dvt = \frac{d}{v}
Where:
• tt is the time,
• dd is the distance,
• vv is the speed of light in the medium.
Since the speed of light in any medium is related to the speed of light in vacuum (cc) by the refractive index μ\mu, we can write:
v=cμv = \frac{c}{\mu}
Thus, the time taken to travel a distance dd in a medium with refractive index μ\mu is:
t=dcμ=μdct = \frac{d}{\frac{c}{\mu}} = \frac{\mu d}{c}
Time taken in glass (t1t_1) and water (t2t_2):
• For glass: The time t1t_1 is given by:
t1=μglass⋅dc=1.5⋅0.093×108t_1 = \frac{\mu_{\text{glass}} \cdot d}{c} = \frac{1.5 \cdot 0.09}{3 \times 10^8} t1=0.1353×108=4.5×10−10 secondst_1 = \frac{0.135}{3 \times 10^8} = 4.5 \times 10^{-10} \, \text{seconds}
• For water: The time t2t_2 is given by:
t2=μwater⋅dc=43⋅0.093×108t_2 = \frac{\mu_{\text{water}} \cdot d}{c} = \frac{\frac{4}{3} \cdot 0.09}{3 \times 10^8} t2=0.123×108=4.0×10−10 secondst_2 = \frac{0.12}{3 \times 10^8} = 4.0 \times 10^{-10} \, \text{seconds}
Difference in time (Δt\Delta t):
Now, we calculate the difference in time:
Δt=t1−t2=4.5×10−10−4.0×10−10=5×10−11 seconds\Delta t = t_1 - t_2 = 4.5 \times 10^{-10} - 4.0 \times 10^{-10} = 5 \times 10^{-11} \, \text{seconds}
Conclusion:
The value of (t1−t2)(t_1 - t_2) is 5 × 10⁻¹¹ seconds.
Thus, the correct answer is: A) 5×10−115 \times 10^{-11} sec.
