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If a diver who is underwater shines a flashlight upward toward the surface, at an angle of 29° from the normal, at what angle does the light emerge from the water? The indices of refraction are: water,1.33 ; air,1.00029

Short Answer

Expert verified

The angle at which the light emerges from the water is 21.38°.

Step by step solution

01

Identification of given data

Refractive index of water,n2=1.33

Refractive index of air,n1=1.00029

The angle of incident,θ1=29°

02

Determine the formulas to calculate the angle at which the light emerges from the water.

The Snell’s law defines the relationship between the angle of incident and angle of refraction when light of wave passes through two mediums.

n1sinθ1=n2sinθ2 …(¾±)

Here,θ2 is the angle of refection.

03

Determining the angle at which the light emerges from the water

Calculate the angle at which the light emerges from water.

Substitute 29°forθ1, 1.33forn2and 1.00029forn1into equation (i)

1.00029sin29°=1.33sinθ2sinθ2=1.000291.33sin29°sinθ2=0.752×0.4848sinθ2=0.3645

Solve further as,

θ2=sin-10.3645θ2=21.38°

Hence the angle at which the light emerges from the water is 21.38°.

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