A diverging lens of focal length -12cm projects the image of an object onto a wall. What is the object distance if the image is virtual, upright, and 45% of the object size?

Question

aakashsudhakar · Answer

Geometric optics... the absolute bane of my AP Physics experience. XD Still, my duty to help!

f = -12, which means that the focal point (and projected image) is on the same side as the object! This means that the object is going to be virtual, but of course, we already know that.

An image 45% of the object means that the magnification of the object, depicted by $$m = -i/o$$ is equal to 0.45. One critical fact that you can infer from this is the fact that either the image or the object distance is negative, since the magnification is positive.

-i/o = 0.45

Given (by Geometric Optics formulae): $$o/i = f/(i - f)$$
$$(o/i)/-1 = -i/o$$
Therefore: $$0.45 = (f/(i - f))/-1$$
$$0.45 = (f - i)/f$$
$$0.45 = 1 - (i/f)$$
$$0.45 = 1 - (i/-12)$$
$$(i/-12) = 0.55$$
$$i = -6.6$$

Now that we know that the image is located -6.6 cm away from the lens center, we can solve for the object distance from the lens center using the Thin-Lens Equation.

$$\frac{ 1 }{ f } = \frac{ 1 }{ i } + \frac{ 1 }{ o }$$
Plug in the known information and solve for o.

$$\frac{ 1 }{ -12 } = \frac{ 1 }{ -6.6 } + \frac{ 1 }{ o }$$
$$o = 1.8181818$$

We have now calculated that the object distance equals ≈1.818182 cm.

anonymous · Answer

Thx for the explanation. Doing this is such a pain and you have explained very well. Thank you :D

aakashsudhakar · Answer

No problem, and tell me about it... you in AP Physics? I just finished the unit on Geometric Optics and Light Mechanics.

anonymous · Answer

Yes i am

aakashsudhakar · Answer

Thankfully it get's a little more fun after this. Just started Electrostatics, which actually isn't bad. Getting through Optics is a pain but not many more units connect too much with it once it's outta the way.

anonymous · Answer

Ok thank you

aakashsudhakar · Answer

Anytime!