In the figure below, suppose that the angle of incidence is θ1 = 29.7°, the thickness of the glass pane is 5.46 mm, and the refractive index of the glass is n2 = 1.52. Find the amount (in mm) by which the emergent ray is displaced relative to the incident ray.

Well, well, well, we have a figure that's got us all geometry enthusiasts excited! Now, let's see, we've got an angle, a thickness, and a refractive index, but what we really need is a punchline! Alright, here goes nothing:

Why did the emergent ray go to art school?

Because it wanted to DISPLACEdly express itself! 😄

But in all seriousness, let's get calculating. To find the displacement of the emergent ray relative to the incident ray, we can use a bit of trigonometry and Snell's Law.

The equation we need is:
displacement = thickness * (tan(θ1) - tan(θ2))

Now, we know the angle of incidence θ1 is 29.7°, but we need to find θ2 using Snell's Law. Snell's Law states:
n1 * sin(θ1) = n2 * sin(θ2)

Given that the refractive index of air (n1) is about 1.00, we can plug in our values:
1.00 * sin(29.7°) = 1.52 * sin(θ2)

Solving for θ2, we find:
θ2 ≈ sin^(-1)((1.00 * sin(29.7°)) / 1.52)

Once we have that, we can plug it back into our displacement equation, using the given thickness of the glass pane (5.46 mm). And voila! You'll have your answer.

Remember, math is no joke, but I like to add a pinch of humor to keep things entertaining. Happy calculating, my fellow comedy connoisseur!

To find the amount by which the emergent ray is displaced relative to the incident ray, we can use the formula:

Displacement = (Thickness of the glass pane) x tan(θ2)

First, we need to find the angle of refraction (θ2) using Snell's law:

n1 * sin(θ1) = n2 * sin(θ2)

Given:
θ1 = 29.7°
n2 = 1.52

Assuming the incident medium is air, which has a refractive index of approximately 1, we can substitute the values into the equation and solve for θ2:

1 * sin(29.7°) = 1.52 * sin(θ2)

sin(θ2) = (1 * sin(29.7°)) / 1.52
θ2 ≈ sin^(-1)((1 * sin(29.7°)) / 1.52)

Using a calculator, we find that θ2 ≈ 19.02°

Now, we can calculate the displacement using the thickness of the glass pane and the tangent of θ2:

Displacement = (5.46 mm) x tan(19.02°)

Using a calculator, we find that the displacement is approximately 1.92 mm.

Therefore, the emergent ray is displaced relative to the incident ray by approximately 1.92 mm.

To find the amount by which the emergent ray is displaced relative to the incident ray, we need to use the concepts of refraction and Snell's law.

Snell's law 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 refractive indices of the two media:

n1 * sin(θ1) = n2 * sin(θ2)

Where:
- n1 is the refractive index of the medium from which the light is coming (in this case, air)
- θ1 is the angle of incidence
- n2 is the refractive index of the medium into which the light is entering (in this case, glass)
- θ2 is the angle of refraction

Now, let's solve for θ2:
sin(θ2) = (n1 / n2) * sin(θ1)
θ2 = arcsin((n1 / n2) * sin(θ1))

To find the displacement of the emergent ray, we need to find the horizontal distance traveled by the ray in the glass pane. This can be calculated using the formula:

Displacement = Thickness * tan(θ2)

Substituting the given values, we have:
Thickness = 5.46 mm (given)
θ1 = 29.7° (given)
n2 = 1.52 (given)
n1 = 1 (refractive index of air)

First, let's calculate θ2:
θ2 = arcsin((n1 / n2) * sin(θ1))
= arcsin((1 / 1.52) * sin(29.7°))
≈ arcsin(0.6657)
≈ 41.4°

Now, let's calculate the displacement using the formula:
Displacement = Thickness * tan(θ2)
= 5.46 mm * tan(41.4°)

Calculating the above expression, we find the displacement of the emergent ray relative to the incident ray.