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.