Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is Ā 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
nāsinĪøā = nāsinĪøā
y = 2.2 m and strikes at x = 8.5 m, therefore tanĪøā = 2.2/8.5 = 0.259 and
Īøā = Ā 14.511 Ā°
nā = 1.0003 = refractive index of air
nā = 1.33 = refractive index of water
Therefore sinĪøā = Ā [tex]\frac{n_1sin\theta_1}{n_2}[/tex] Ā = [tex]\frac{1.003*0.251}{1.33}[/tex] = 0.1885 and Īøā = 10.86 Ā°
Since the water depth is 4.0 m we have tanĪøā = [tex]\frac{4}{x_2}[/tex] or xā = [tex]\frac{4}{tan\theta_2 }[/tex] =[tex]\frac{4}{tan(10.86)}[/tex] = 20.845 m
d = xā + 8.5 = 20.845 m + 8.5 m = 29.345 m.
