2
$\begingroup$

In this particular derivation of refraction happening at a spherical surface in terms of its radius of curvature , image and object distance and refractive index is done by my book as shown

enter image description here

enter image description here

When we look at equation 3 and 4 , we apply sign convention at 3 because in this case U is negative. But a numerical question just after this is: Takes U to be negative again , isn't that already done before when we derived lens formula , why is it that this particular case , takes U to be negative rather since we have derived a formula for this exact instant which already accounts for the negative?

enter image description here

$\endgroup$

1 Answer 1

2
$\begingroup$

Formula $(3)$ works for the situation shown above.

If $O$ is brought closer to the lens then the ray $NI$ diverges away from the principal axis and when back produced intersect the principal axis to the left of the interface. On performing a similar analysis the second term in $(3)$ has a negative sign in front of it.
One finds that thing change if the interface has the opposite curvature.
Thus, the form of formula $(3)$ varies as per situation.

The introduction of a sign convention means the the same formula $(4)$ can be used for all situations.

The justification for the step $OM=-u,\,MI=+v,\,MC=+R$ is that it produces a formula $(4)$ which, if the sign convention is applied correctly, works for all situations. That is it.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.