TBR optics p286 please help!

[yellowjellybean]

---

Members don't see this ad.

Passage 7 on the optics section, p 286

When an object is placed inside the focal length on the right side of a diverging lens, what type of image is formed?

I know the answer is " virtual, upright image" conceptually.

But the explanation is driving me crazy!!!!

Since the object is to the right of the lens and the lens is diverging, both p and q are negative. If the object is placed inside the focal length, the magnitude of f is greater than the magnitude of p, so the magnitude of 1/p is greater than the magnitude of 1/f. Since 1/p is a larger negative number, then in order to satisfy the lens equation, q must be positive. If q is positive, the magnification equations indicates that the image is upright.

How can the bold above both be true?

 

[jakernewt]

It should say "p and f are negative" in the first bolded statement.

 

[yellowjellybean]

Thank you for the reply! But I am still confused- how is it that q comes out positive? TBR says if q>0 then the image is real. But we know the image is virtual because this is a diverging lens. I'm just trying to understand this +/- system.

 

[ShySpliceosome]

I'm bumping this thread, because I too am confused about how q can be both positive and negative 😕 ...it just doesn't make sense...

 

[FeinMS]

I'm bumping this thread, because I too am confused about how q can be both positive and negative 😕 ...it just doesn't make sense...

As it says above, it's a typo and first bolded sentence should say "p and f are negative."