Question: (a) A tank containing methanol has walls 2.50 cm thick made
of glass of refractive index...
Question
(a) A tank containing methanol has walls 2.50 cm thick made
of glass of refractive index...
(a) A tank containing methanol has walls 2.50 cm thick made
of glass of refractive index 1.550. Light from the outside air
strikes the glass at 41.3 degrees angle with the normal to the
glass. Find the angle the light makes with the normal in the
methanol.
(b) The tank is emptied and refilled with an unknown liquid.
If light incident at the same angle as in part (a) enters the
liquid in the tank at an angle of 20.2 from the normal, what is the
refractive index of the unknown liquid?
The concept required to solve the given problem is Snell’s law of refraction.
Initially calculate the angle the light makes with the normal in the methanol and later calculate the refractive index of the unknown liquid will be calculated.
Fundamentals
Snell’s law of refraction states that the ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given color and for the given pair of media.
The expression for the Snell’s law of refraction is,
μ1sin(i)=μ2sin(r) …… (1)
Here, μ1 is the refractive index of air, μ2 is the refractive index of glass, i is the angle of incidence, and r is the angle of refraction.
(a)
The expression to calculate angle is,
r=sin−1(μ2μ1sin(i))
Substitute 41.3∘ for i, 1 for μ1 and 1.55 for μ2 in the above equation.
r=sin−1(1.55sin(41.3∘))=1.55
Substitute 25.2∘ for i, 1.55 for μ1 , and 1.329 for μ2 in equation (1).
r=sin−1(1.3281.55sin(25.2∘))=29.8∘
(b)
The refractive index of an unknown liquid is,
μ1sin(i)=μ2sin(r)
Substitute 1.55 for μ1 , 25.2∘ for i, and 20.2∘ for r in the above equation.