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Lesson 3: Lenses

# Object image and focal distance relationship (proof of formula)

The video provides a detailed walkthrough of deriving the lens formula, which relates the object distance, image distance, and focal length in a convex lens. It uses similar triangles and algebraic manipulation to arrive at the formula $1/f = 1/d_0 + 1/d_i$, a key concept in optics for premed students. Created by Sal Khan.

## Want to join the conversation?

• I have learnt the formula of focal length (1/f = 1/di-1/do).Is it wrong?
• no the formula 1/f=1/di- 1/do is correct, because it is based on the sign covention for lenses where object distance (do) is always taken as negative for all real objects, but sal's formula is also not wrong because he did not stick to the sign convention ( he took the object distance as positive)
for eg -
for a covex lens, if f=10cm do=15 di=X
according to ur formula (1/f=1/di-1/do) which is based on sign convention
1/f =1/di -1/do
f is positive(10)
do is negative (-15)
so , 1/10 =1/di - (-1/15)
1/10= 1/di + 1/15
1/di= 1/10-1/15 =1/30
take the reciprocal of i/di
di=30 cm (it is positive)
now we take salman's formula
1/f= 1/di +1/do (remember we are not taking sign conventions we are simply putting the values)
1/10= 1/di +1/15 (not applying sign convention)
1/di=1/10 -1/15 =1/30
we take the reciprocal of 1/di
and di = 30 cm
thus both the formulas are correct ! :)
• Why does he keep saying D-NOT while writing Do when the subtitles say D0?
• He's saying d-naught. Naught is another term for nothing or zero "0".
• Is this formula applicable to convex mirrors? If it is then can you do a video on its proof
• No, the lens was a convex lens but for this to be same we need to have a concave mirror
• How can I solve this formula? Equation for focal point: formula: 1/u +1/v= 1/f
• If you mean solving for a particular variable, here you go:
1/u + 1/v = 1/f
f/u + f/v = 1
f + fu/v = u
fv + fu = uv
Now that there aren't any confusing denominators, we can solve for any variable:
fv + fu = uv
f(v + u) = uv
f = uv / (v + u)

fu + fv = uv
fv = uv - fu
fv = u (v - f)
u = fv / (v - f)

Hope this helps!
• What's a "Focal Plane"?
• A plane passing through the principal focus and at right angles to the principal axis of a spherical mirror is called the focal plane.
• Does this equation work with convex and concave mirrors as well?
• There is a difference between solving a convex lens and a concave lens
• if the object is between the lens and second focal length...how would the image be??
• In this case, an erect virtual image of the object would be formed to the right of the object. The image would also be bigger than the object.

However, this isn't the conventional way of imagining the way you would place the lens and the object. After placing the object between the lens and the second focal length, you could view it from the other side, as if you turned the table around. Now the object seems as if it's placed between the first focus and the lens. Then you can proceed with doing it like it's always done!

The first and second focus are actually equal. You can obtain their value using the Lens Maker's Formula (I hope I'm not wrong. If I am, someone please correct me).
• FOCAL LENGTH...NEVER CONSTANT!
According to our science textbook focal length(f) of a mirror/ lens is equal to half its radius of curvature(R).Which means that there is a specific focal length for a given lens/mirror(f=R/2).
But in science lab when we perform the experiment to find the focal length of a lens/mirror we get different focal lengths for the same mirror/lens for different objects(that we observe)...this means that the focal length of a mirror/lens is never constant...so the formula f=R/2 is contradicted!
In fact...what i think is that the ray of light parallel to principle axis never pass through the focus[what our science textbook states (f=R/2)] after reflection/refraction!
is it correct?
• 1) real lenses do not exactly follow the formula that we devise for an ideal lens. Real lenses never produce perfectly focused images
2) possibly, you are seeing some diffraction.
3) if you are not using monochromatic light then the focal length is not exactly the same for all wavelengths