A laser beam is incident at an angle of 30.2° to the vertical onto a solution of corn syrup in water. (a) If the beam is refracted to 18.82°

Question

A laser beam is incident at an angle of 30.2° to the vertical onto a solution of corn syrup in water. (a) If the beam is refracted to 18.82° to the vertical, what is the index of refraction of the syrup solution? (b) Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its wavelength in the solution. nm (c) Find its frequency in the solution. Hz (d) Find its speed in the solution.

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Farah 4 years 2021-07-19T07:16:38+00:00 1 Answers 48 views 0

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  1. Doris
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    2021-07-19T07:17:50+00:00
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    Answer:

    a) n2 = 1.55

    b) 408.25 nm

    c) 4.74*10^14 Hz

    d) 1.93*10^8 m/s

    Explanation:

    a) To find the index of refraction of the syrup solution you use the Snell’s law:

    n_1sin\theta_1=n_2sin\theta_2   (1)

    n1: index of refraction of air

    n2: index of syrup solution

    angle1: incidence angle

    angle2: refraction angle

    You replace the values of the parameter in (1) and calculate n2:

    n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55

    b) To fond the wavelength in the solution you use:

    \frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm

    c) The frequency of the wave in the solution is:

    v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz

    d) The speed in the solution is given by:

    v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s

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