Answer:
Angle of ray makes with the vertical is 62.1 degree
Explanation:
As per the ray diagram we know that the angle of incidence on oil brine interface will be given as
now by Snell'a law at that interface we have
now we will have
now this is the angle of incidence for oil air interface
so now again by Snell's law we will have
Answer:
5.2 m
Explanation:
from the question we are given the following
depth of pool (d) = 3.2 m
height of laser above the pool (h) = 1 m
point of entry of laser beam from edge of water (l) = 2.5 m
we first have to calculate the angle at which the laser beam enters the water (∝),
tan ∝ = \frac{1}{2.2}
∝ = 24.44 degrees
from the attached diagram, the angle with the normal (i) = 90 - 24.4 = 65.56 degrees
lets assume it is a red laser which has a refractive index of 1.331 in water, and with this we can find the angle of refraction (r) using the formula below
refractive index = \frac{sin i}{sin r}
1.331 = \frac{sin 65.56}{sin r}
r = 43.16 degrees
we can get the distance (x) from tan r = \frac{x}{3.2}
tan 43.16 = \frac{x}{3.2}
x = 3 m
To get the total distance we need to add the value of x to 2.2 m
total distance = 3 + 2.2 = 5.2 m
