Aamc 11 #23

[cfive22]

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I'm very confused as to the reasoning for #23.

The optical power P of a lens is expressed in diopters: P=1/f, where f is the focal length in meters. Given that the lens-to-retina distance in the human eye averages 2.0 cm, which of the following is closest to the power of the lens of the human eye?

A)0.05
B)2
C)10
D)40

Answer: D. The lens power P of the eye is one over the focal length 1/f. The focal length of the unaided eye is 2 cm = 2 x 10^-2 m; therefore, the power is P = 1/(2 x 10^-2 m) = 50 diopters. Option D, 40, is the closest answer.

I get the reasoning and I got the question right. However, just a question on lenses. If the focal length of a converging lens is the point at which the light converges, what is the radius of curvature? I guess I'm a bit confused as to the difference in this aspect between converging mirrors and lenses.

 

[MedPR]

I'm very confused as to the reasoning for #23.

The optical power P of a lens is expressed in diopters: P=1/f, where f is the focal length in meters. Given that the lens-to-retina distance in the human eye averages 2.0 cm, which of the following is closest to the power of the lens of the human eye?

A)0.05
B)2
C)10
D)40

Answer: D. The lens power P of the eye is one over the focal length 1/f. The focal length of the unaided eye is 2 cm = 2 x 10^-2 m; therefore, the power is P = 1/(2 x 10^-2 m) = 50 diopters. Option D, 40, is the closest answer.

I get the reasoning and I got the question right. However, just a question on lenses. If the focal length of a converging lens is the point at which the light converges, what is the radius of curvature? I guess I'm a bit confused as to the difference in this aspect between converging mirrors and lenses.

Isn't the radius of curvature always 2f? I spent a little extra time on this question because 50 just wasn't close enough to 40 for me.

 

[sudo]

I don't know why the hell they wouldn't just have 50 as an answer choice. It's not like it's a decimal answer! ...I guess the test writers can't stand giving freebie questions too easily.

Sent from my Nexus S 4G using Tapatalk

 

[cfive22]

Isn't the radius of curvature always 2f? I spent a little extra time on this question because 50 just wasn't close enough to 40 for me.

I always thought that the radius of curvature for a lens or mirror was pretty much the diameter of the mirror or lens if it were to make a full circle and the focal length would be half that value. I guess that reasoning is only good for mirrors and lens you just double the focal length, which is where the light rays converge.

 

[MedPR]

I don't know why the hell they wouldn't just have 50 as an answer choice. It's not like it's a decimal answer! ...I guess the test writers can't stand giving freebie questions too easily.

Sent from my Nexus S 4G using Tapatalk

Yup, this is exactly why I paused after doing the calculation.

I always thought that the radius of curvature for a lens or mirror was pretty much the diameter of the mirror or lens if it were to make a full circle and the focal length would be half that value. I guess that reasoning is only good for mirrors and lens you just double the focal length, which is where the light rays converge.

Yea, I don't think we'll get any exceptions on the MCAT.

 

[dipbk]

A perhaps silly question, but how do we know that the focal length is 2cm? I thought that distance from lens to retina would be the image distance, (1/s in the thin-lens equation)....

 

[dnrs]

I believe it's because the image formed at the focal point is clearest.