Question

A ray of light starts at ((8, 4)) and travels through a pinhole at ((0, 0)). It continues until it hits an image plane 666 units behind the pinhole at a point ((x₁, y₁)).

Where does the ray hit the image plane?

a) At (x₁, y₁)

b) At (0, 666)

c) At (8, 670)

d) At (666, 0)

Answer 1

The ray hits the image plane at (-666, -333).

Step-by-step explanation:

In order to find where the ray hits the image plane, we need to trace the path of the ray. Since the ray starts at (8, 4) and travels through a pinhole at (0, 0), it follows a straight line. The equation of the line can be found using the slope-intercept form: y = mx + b. The slope (m) can be calculated as (change in y)/(change in x) = (4-0)/(8-0) = 1/2. Plugging in the values, the equation becomes: y = (1/2)x + b. To find the value of b, we can substitute one of the given points (8, 4) into the equation: 4 = (1/2)*8 + b. Solving for b, we get: b = 4 - 4 =0. Therefore, the equation of the line is: y = (1/2)x + 0.

The image plane is located 666 units behind the pinhole. This means that the x-coordinate of the image point will be 666 units behind the x-coordinate of the pinhole. Since the pinhole is at (0, 0), the image point will be at (-666, y).

Substituting this value into the equation of the line, we get: y = (1/2)*(-666) + 0. Solving for y, we get: y = -333. Therefore, the ray hits the image plane at (-666, -333).