An object is placed on the end (0 cm) of a two-meter stick. At the 150 cm mark i

Question

An object is placed on the end (0 cm) of a two-meter stick. At the 150 cm mark is placed a screen for viewing images. You are supplied with a converging ( f = 22.5 cm) lens.

a) Find the two positions on the two-meter stick where you can place the lens so that a sharp image of the object will appear on the screen. For each solution, calculate the magnification M.

b) Construct a ray diagram for each solution.

Light enters the 6.0 cm wide square prism above at the midpoint of one side.

At what angle should you shine an incident ray so that the refracted ray hits the corner prism?

Estimate the location where the ray leaves the prism if the incident angle is half as large as in part a).

Explanation / Answer

Number 1)

Apply 1/f = 1/p + 1/q

q + p = 150 cm, so q = 150-p

1/22.5 = 1/p + 1/(150-p)

1/22.5 = (150-p + p)/(150 - p)(p) -- Cross Multiply

150p - p2 = 3375 - Put in standard form

p2 - 150p + 3375 = 0

Find the two roots in the Quadratic Equation

The roots are...122.4cm and 27.6 cm

So place the lens at either 27.6 cm or at 122.4 cm

The magnification at 27.6...

M = -122.4/27.6 = -4.34 (Negative means inverted)

At 122.4, magnification is...

M = -27.6/122.4 = -.225 (Negative means inverted)

Number 2)

To hit the bottom corner, we need the refracted angle by the geometry of the square

tan(angle) = 3/6

angle = 26.6o

Then apply Snells Law

n1sin(i) = (n2)sin (r)

1(sin i) = 1.45(sin 26.6)

i = 40.4o

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