Power of a thin lens in a medium is 1/fₘ or μₘ/fₘ ?

[shubhamakshit]

Homework Statement

A thin glass (refractive index 1.5) lens has optical power of-5 D in air. lts opticat power in a liquid medium with refractive index 1.6 will be

Relevant Equations

P = 1/fₘ = (μₗₑₙₛ₋ₘ - 1)[ 1/R₁ - 1/R₂]

Here is what I tried

This question was actually asked in one of our engineering entrances.

The answer was 1D.

My teachers say that we have to use μₘ/fₘ to get to this answer. I cannot understand why. I'll be really glad if you could tell me the exact definition of power (numerically) that works in all scenarios.

 

[Steve4Physics]

Relevant Equations: P = 1/fₘ = (μₗₑₙₛ₋ₘ - 1)[ 1/R₁ - 1/R₂]

Welcome to PF @shubhamakshit. The above formula only applies to a thin lens when the surrounding medium is air/vacuum (refractive index = 1).

The power of a lens in a medium, m, is the refractive index of the medium divided by the focal length in the medium: $P_m = \frac {\mu_m}{f_m}$. When the medium is air/vacuum this simplifies to $P = \frac 1f$.

The general formula is $P_m = \frac {\mu_m}{f_m} = (\mu_{lens} - \mu_m) [\frac 1{R_1} - \frac 1{R_2}]$.

See if you can get the correct answer using that information.

 

[shubhamakshit]

Well that did give me the answer.
Thanks a lot.

I was still not happy as to why the 'power' which I feel is a property of lens alone must depend on
μ in addition to the fact that power depends on focal length which itself depends on μ.

Power is essentially a measure of how much a lens can bend a light ray.

However I finally reasoned that lets say we have two conditions

A lens which has lets say +20cm focal length in air - A
Another lens of same material which has also +20cm focal length in water - B

If I go by my initial thought (P=1/f) we would have force to conclude that bending power of A is same as that of B which would be a wrong statement.

I'll be really happy if you guide me by explaining if my explanation is correct.
Thanks

 

[Steve4Physics]

I was still not happy as to why the 'power' which I feel is a property of lens alone must depend on
μ in addition to the fact that power depends on focal length which itself depends on μ.

Power is essentially a measure of how much a lens can bend a light ray.

However I finally reasoned that lets say we have two conditions

A lens which has lets say +20cm focal length in air - A
Another lens of same material which has also +20cm focal length in water - B

If I go by my initial thought (P=1/f) we would have force to conclude that bending power of A is same as that of B which would be a wrong statement.

I'll be really happy if you guide me by explaining if my explanation is correct.
Thanks

I’m not entirely sure why optical power is defined as $P_m = \frac {\mu_m}{f_m}$.

I suspect that the main reason is to enable powers to be added where more than one medium is present. For example, if you want the overall power for a lens with one surface in air and the other surface in a liquid. (The power of a thin lens is approximately the sum of the power of the surfaces.)

 

[shubhamakshit]

Thank you all the same for I got my answer from the general formula you provided.

 

[NTesla]

Welcome to PF @shubhamakshit. The above formula only applies to a thin lens when the surrounding medium is air/vacuum (refractive index = 1).

The power of a lens in a medium, m, is the refractive index of the medium divided by the focal length in the medium: $P_m = \frac {\mu_m}{f_m}$. When the medium is air/vacuum this simplifies to $P = \frac 1f$.

The general formula is $P_m = \frac {\mu_m}{f_m} = (\mu_{lens} - \mu_m) [\frac 1{R_1} - \frac 1{R_2}]$.

See if you can get the correct answer using that information.

I'm not sure where you got this formula from. I've referred to more than 1 book, but couldn't find your version of the formula. I've gone through the derivation of lens maker's formula, but couldn't find any reason, why the power should be defined the way you have written.

I think, the initial calculation by OP is correct. P = 5/8.

 

[NTesla]

I’m not entirely sure why optical power is defined as $P_m = \frac {\mu_m}{f_m}$.

I suspect that the main reason is to enable powers to be added where more than one medium is present. For example, if you want the overall power for a lens with one surface in air and the other surface in a liquid. (The power of a thin lens is approximately the sum of the power of the surfaces.)

The derivation of lens maker's formula already encapsulates the condition of refractive index of medium being different from that of the material of the lens. There's no need to further include the refractive index of the medium while defining power of the lens.

 

[shubhamakshit]

The derivation of lens maker's formula already encapsulates the condition of refractive index of medium being different from that of the material of the lens. There's no need to further include the refractive index of the medium while defining power of the lens.

That's what I initially thought,
But my teacher as well as the solution to the question asked in the entrance all seemed to disagree.

But I satisfied myself by the argument I gave here https://www.physicsforums.com/threa...-a-medium-is-1-f-or-m-f.1082649/#post-7285050

 

[NTesla]

Well that did give me the answer.
Thanks a lot.

I was still not happy as to why the 'power' which I feel is a property of lens alone must depend on
μ in addition to the fact that power depends on focal length which itself depends on μ.

Power is essentially a measure of how much a lens can bend a light ray.

However I finally reasoned that lets say we have two conditions

A lens which has lets say +20cm focal length in air - A
Another lens of same material which has also +20cm focal length in water - B

If I go by my initial thought (P=1/f) we would have force to conclude that bending power of A is same as that of B which would be a wrong statement.

I'll be really happy if you guide me by explaining if my explanation is correct.
Thanks

Your reasoning is not correct. In both the cases, A and B, the power of the lens is same. We would be wrong to say that "Both the lens have same power" unless we specify whether both the lenses are in same medium or not.

 

[NTesla]

But my teacher as well as the solution to the question asked in the entrance all seemed to disagree.

Did your teacher give any reasoning thereof ?

 

[shubhamakshit]

Well no...
And now that I think of it I am confused once again.

 

[NTesla]

Well no...
And now that I think of it I am confused once again.

Well, confusion is the most fertile state of mind. It means that you want to understand something deeply. Don't ever accept something just because of authority of the person. Question everything.

 

[Steve4Physics]

As I understand it, NEET (UG) is an entrance examination set by the Indian National Testing Agency (NTA) for entrance to medical degrees,

Study-material is provided by the NTA. This is a link to the study material covering ‘Power of a Lens’: https://unacademy.com/content/neet-ug/study-material/physics/power-of-a-lens-2/

Edit: The link is not from the NTA. Unacademy is an independent company and its material is not authoritative.

Here is a quote from the link:

“Power of lens in a medium
.
Power of the lens is equal to the refractive index of any medium other than air which is divided by its focal length.
P = n / f….equation (1). “

I can’t understand the rationale behind this definition. Conceivably it’s a mistake.

So, although dubious, that’s the definition used for NEET (UG) and possibly for other examinations.

Edit.

 

[NTesla]

Study-material is provided by the NTA

I don't understand how did you reach that conclusion.
Study material is NOT provided by NTA. As far as I know, only syllabus is provided by NTA in addition to conducting the exam.
That link that you've shared is not an authentic source. It's just something written by someone on internet, without offering any proof or rationale whatsoever, for the point under consideration. That doesn't make the statement right.

 

[Steve4Physics]

I don't understand how did you reach that conclusion.
Study material is NOT provided by NTA. As far as I know, only syllabus is provided by NTA in addition to conducting the exam.
That unacademy link that you've shared is not an authentic source. It's just something written by someone on internet, without offering any proof or rationale whatsoever. That doesn't make the statement right.

Yes, you’re right. I had (badly) misunderstood the nature of unacademy. It is not linked to NTA. It appears to be a (fairly large, established) educational company - but definitely not an authoritative source. Apologies for the confusion.

I had found this earlier:
https://physics.stackexchange.com/questions/810635/formula-for-power-of-a-lens-submerged-in-a-medium
which resolved the OP's problem. That's what prompted my original reply.

As already noted, I can’t see any rationale to define power as $\frac {\mu}f$. So it looks like a mistake which has somehow survived and propagated in a limited environment.

I’m not sure what the OP should do if encountering a question (such as the Post #1 question) in a formal exam.

Minor edit.