What is the maximum radius that this circle could have?

[gamesandmore]

A point source of light is submerged 2.6 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have?

No idea where to begin this one - nair = 1.00 nwater = 1.33 and
n1sin01 = n2sin02

 

[Doc Al]

Draw yourself a picture showing various rays of light emitted in all directions. What happens as they hit the surface of the water? Examine how the angle of incidence relates to the angle of refraction.

 

[ogb p]

The maximum radius that this circle could have can be determined by 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 indices of refraction of the two media. In this case, the media are air and water, with indices of refraction of 1.00 and 1.33 respectively.

Since the point source of light is submerged 2.6 m below the surface of the lake, the angle of incidence (01) can be calculated using the equation sin01 = h/D, where h is the depth of the source and D is the distance from the source to the surface. In this case, h = 2.6 m and D = 2.6 m, so sin01 = 2.6/2.6 = 1.

Using this value for the angle of incidence, we can rearrange Snell's Law to solve for the angle of refraction (02). The equation becomes n1sin01 = n2sin02, where n1 is the index of refraction of air and n2 is the index of refraction of water. Substituting in the values, we get 1.00 x 1 = 1.33 x sin02. Solving for sin02, we get sin02 = 0.75.

Now, using the equation for the area of a circle (A = πr^2), we can calculate the maximum radius (r) of the circle of light on the surface of the lake. The angle of refraction (02) can be used as the angle of the sector (in radians) in the equation. Thus, the maximum radius can be calculated as r = D x sin02 = 2.6 x 0.75 = 1.95 m.

Therefore, the maximum radius of the circle of light on the surface of the lake is 1.95 m. This means that the area illuminated on the surface of the lake is a circle with a maximum radius of 1.95 m, with the point source of light submerged 2.6 m below the surface.