2
$\begingroup$

Let's say that a concave lens $A$ has a focal length of $-10$ $cm$ and another concave lens $B$ has a focal length of $-20$ $cm$ and we need to compare their powers.

We know that : $$P(D) = \dfrac{1}{f(m)}$$ Here, $P(D)$ means the power of the lens in dioptres (D) and $f(m)$ means the focal length of the lens in metres (m).

Here, $f_A=-10$ $cm$ and $f_B = -20$ $cm$.
So, $P_A = -10D$ and $P_B = -5D$.

Now, mathematically thinking, $-5 > -10$, so $P_B > P_A$ and hence $B$ should have a greater power.
But, logically thinking, lens $B$ has a focal length of $-10$ $cm$ and lens $B$ has a focal length of $-20$ $cm$. So, rays parallel to the principal axis will seem to meet $10$ $cm$ from the optical centre upon refraction from lens $A$ and $20$ $cm$ from lens $B$. So, the rays are diverging more in the case of lens $A$ since they appear to meet closer to the optical centre.

So, in my opinion, lens $A$ is more powerful than lens $A$ and we should compare the powers of concave lens by looking at the absolute values of their focal lengths and powers when comparing and negative values for magnitude and nature of the lens.
For example, $$|P_A| = 10D \text { and }|P_B|=5D$$ $$\text {So, }|P_A| > |P_B|$$ Thus, lens $A$ is more powerful than lens $B$

Let me know if I'm right.
Thanks!

$\endgroup$
1
$\begingroup$

Yes, you are right. Compare only the magnitudes of lens powers. Because the power of a lens shows ability of lens to bend the light rays after refraction.

The sign of power of lens shows the nature of lens i.e. converging (convex lens) or diverging (concave lens). The power of two lenses are compared with there magnitude (without sign) i.e. their magnitudes of powers are compared.

$\endgroup$
1
$\begingroup$

Your first step, that mathematically speaking $-5\:\mathrm D>-10\:\mathrm D$, is correct but in lenses that negative sign only represents which type of lens it is. So lens $A$ has greater power than $B$.

$\endgroup$
1
  • $\begingroup$ Thanks, I just saw your question too :P $\endgroup$ May 21 '20 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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