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anonymous

  • 5 years ago

To a person under water, looking upwards, all objects above and outside water appear to be within a certain cone of vision. If the refractive index of water is 1.33, calculate the vertical angle of the cone.

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  1. anonymous
    • 5 years ago
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    Use Snell's law: \[\ n_1sin(\theta_1)=n_2sin(\theta_2)\] or better yet \[\ \frac{sin(\theta_1)}{sin(\theta_2)}=\frac{n_2}{n_1} \] where:|dw:1327490230845:dw| n_1 - is the refractive index of the medium above water (air probably) and its refractive index is close to 1 so take it as n_1 = 1 n_2 is the refracting index of water, n_2 = 1.33 theta_1 - angle of the incoming ray with the normal (in air) theta_2 = angle of the outgoing ray with the normal (in water). You can revert the direction of the lines in my picture and get the situation in your problem, but it doesn't change the result.

  2. JamesJ
    • 5 years ago
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    The question then is: what's the maximum angle \( \theta_2 \). That will occur when \[ \theta_1 = \pi/2 \] and hence \[ \sin(\theta_1) = \sin(\pi/2) = 1 \] Now, by Snell's law \[ \frac{n_1}{n_2} = \frac{\sin \theta_2}{\sin \theta_2} \] Hence \[ \frac{1}{1.33} = \frac{\sin \theta_2}{1} = \sin \theta_2 \] \[ \implies \theta_2 = \arcsin(1.33) = 48.75 \ degrees \] Hence the cone has an angle to the vertical of 48.75 degrees.

  3. JamesJ
    • 5 years ago
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    **correction: should of course be \[ \frac{n_1}{n_2} = \frac{\sin \theta_2}{\sin \theta_1} \] not theta 2 / theta 2.

  4. anonymous
    • 5 years ago
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    Thank you. The answer is correct.

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