Questions about light and lenses?

Question

A 2 cm tall candle is 10.9 cm from a converging lens that has a focal length of 7.40 cm. How far from the lens will the image be, is it real or virtual, is it reduced enlarged or the same size as the object, and what is its height?

Monochromatic light passes through a single slit, .400mm wide. It falls on a screeen 1.6 m away. If the distance from the center of the pattern on the screen to the first band in 2.0 mm, what is the wavelength of the light?

Answer 1

Problem #1

distances from lens: 1/object + 1/image = 1/focal
object distance = 10.9 cm; focal distance = 7.4 cm

1/10.9 + 1/i = 1/7.4
1/i = 1/7.4 - 1/10.9 = 4.34E-2
i = 23 cm >> image distance

Magnification = image distance / object distance
= 23 cm / 10.9 = 2.11
object is 2 cm high, image is 2 * 2.11 = 4.22 cm high and inverted

Problem #2

b = single slit width, 0.400mm
L = slit-to-screen distance, 1.6 m (1600 mm)
y = center-to-1st band distance, 2 mm
λ = wavelength

IF L is many times more than y, then:

λ = by/L = 0.4 * 2.0 /1600 = 5E-05 mm

Answer 2

Gross generality, without a secondary lens the objective generally returns a same size or smaller image.

One would have to do the math on this

We know infinity is 7.4 CM so you'd have to figure out the angle of the objective and using caculus, probably, the shift from

f Inf
to
f 10.9cm
based on the angle of the lens, I would assume.

There is probably a fixed formual to derive the curve or angle of the lens based on the inf focal length.

I'm assuming 7.4 is the INFINITY focal length or the point at which the sun would form a disk on a surface.

I would tend to speculat you are looking in the range of 10-12 CM based on my work with macro lenses.

Now as the crow flies.

I'd optical bench the lens and candle and take an actual measurement.

I'd then move the candle further away twice and take more mesurements

Then I'd see if there was a logarithmic factor I could device between the 7.4 INF and the three measured distances.

But that's cheating.

Answer 3

distances from lens: 1/object + 1/image = 1/focal
object distance = 10.9 cm; focal distance = 7.4 cm

1/10.9 + 1/i = 1/7.4

1/i = 1/7.4 - 1/10.9 = 4.34E-2

i = 23 cm >> image distance

Magnification = image distance / object distance
= 23 cm / 10.9 = 2.11
object is 2 cm high, image is 2 * 2.11 = 4.22 cm high and inverted

Answer 4

Go ahead and cheat, we want an answer here!