Optics - Beam illuminating coin under water

[frozen7]

A coin lies on the bottom of a pool under 1.5 m
of water and 0.90 m from the side wall as shown
in Figure 2. If a light beam is incident on the
water surface at the wall,at what angle relative
to the wall must the beam be directed so it will
illuminate the coin. Ans: 28.90

The answer I get is 43.29, different with the given answer? May I know which answer is correct actually?
Thanks.

 

[lightgrav]

I don't see a diagram ... but if :
the ray's angle in the water (measured from the vertical) is about 31 degrees,
then the ray's angle in air (measured from the vertical) is about 43.18 degrees.
Are they REFLECTING the ray off the wall? (I expected this to be a refraction)

 

[quicknote]

I would first like to clarify that the answer to this question may vary depending on certain assumptions and factors, such as the refractive index of water and the type of light source being used. However, based on the given information, I will provide an explanation for the answer of 28.90 degrees.

To determine the angle at which the light beam must be directed in order to illuminate the coin, we need to consider the principles of optics and refraction. When light travels from one medium to another, such as from air to water, it changes direction due to the difference in the speed of light in each medium. This phenomenon is known as refraction.

In this scenario, the light beam is incident on the water surface at an angle of 90 degrees (perpendicular to the surface) at the wall. As it enters the water, it will bend towards the normal (an imaginary line perpendicular to the surface) and continue to travel at a different angle. The exact angle of refraction can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two media.

In this case, the angle of incidence is 90 degrees and the angle of refraction is the angle at which the light beam will hit the coin. Using the given distances, we can calculate the distance between the wall and the coin as 1.5 m + 0.90 m = 2.4 m. This distance, along with the depth of the water (1.5 m), can be used to calculate the angle of refraction using basic trigonometry.

Using the formula sin θ = opposite/hypotenuse, we get sin θ = 1.5/2.4 = 0.625. Taking the inverse sine of 0.625, we get an angle of 37.38 degrees. However, this is the angle of refraction at the water-air interface. To determine the angle of refraction at the water-coin interface, we need to subtract this angle from 90 degrees, giving us 90 - 37.38 = 52.62 degrees.

Since the angle of incidence at the wall is 90 degrees, the angle of reflection will also be 90 degrees. Using the law of reflection, we can determine that the angle between the incident light