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A light bulb is immersed in a tank of liquid with refractive index 1.52. A ray of light from the bulb is incident on the water/air interface at an angle of 56 from the horizontal interface. > At what angle, in degrees from the normal, does the refracted ray exit into the air? Grade S Deducti Potentia Late We Late Po sin0 tan0 acoso coso asin0 Submiss Attempt atan acotan) sinh0 coshtanhcotanho 0 Degrees O Radians 0 detailed Hint Hints: 4% deduction per hint. Hints rernaining: 2 Feedback: deduction per feedback. % All content C

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Answer:

69.358°

Explanation:

[tex]\theta_2[/tex] = Angle at the interface = [tex]90-52=38^{\circ}[/tex]

[tex]\theta_1[/tex] = Required angle

[tex]n_1[/tex] = Refractive index of air = 1

[tex]n_2[/tex] = Refractive index of liquid = 1.52

From Snell's law we have

[tex]n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow theta_1=sin^{-1}\dfrac{n_2sin\theta_2}{n_1}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1.52\times sin38}{1}\\\Rightarrow \theta_1=69.358^{\circ}[/tex]

The angle from the normal that the refracted ray exit into the air is 69.358°

The refracted ray exits into the air at an angle of 58° from the normal which is explained below.

Snell's Law:

Given information:

Light is incident from the liquid, which is a denser medium, on the liquid-air interface, and will emerge out of the liquid to the air, which is a rarer medium.

the angle of incidence of light, i = 90° - 56° = 34°

the refractive index of liquid, n₁ = 1.52

the refractive index of air, n₂ = 1

let the angle of refraction of light be r.

According to Snell's Law:

[tex]\frac{sin(i)}{sin(r)}=\frac{n_2}{n_1} \\\\r=sin^{-1}[\frac{n_1}{n_2}sin(i)] \\\\r=sin^{-1}[1.52\times sin34]\\\\r=sin^{-1}[0.85][/tex]

r = 58°

Learn more about Snell's Law:

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