TPR Physics WB Passage 44 Question 4

[KoalaT]

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Its an optics passage, so I'll do must best to make a diagram here.

Mirror M....................................................Partial Mirror P (slanted)
|MM|-------->-------------------------[W]-------<-----\MP\----->-----[Eye]..._
...............................................................................................^..................................|
...............................................................................................^..................................a
..........................................................................................[light]............................. |
.
|<-------------------L------------------->|<-------a------->|<-------a------>|

**The dots are just place fillers. Look at the bolded dashes for the apparatus

So there is Mirror M which is a plane mirror. Mirror P is a partial mirror and slanted toward Mirror M. W is a wheel with slits, which is unimportant for my specific question. The light source comes from below MP, bounces to MM, and travels back through MP to the eye. The vertical distance from the eye to the light source is a, horizontal distance from eye to light is a, and the distance from the wheel to MM is L.

Q4: Seen through the eyepiece, the light source will appear to be located at a distance of:
A. L- 2a in front of Mirror M.
B. L-a behind Mirror M.
C. L + a behind Mirror M.
D. L + 2a behind Mirror M.

Okay, I put D. The answer is completely in the relation to the mirror M, so I started there. If you travel to the light source from Mirror M you go across "L", "a", and then the vertical distance "a" to the light source. Thus, L + 2a behind the mirror because the image MM is seeing from MP is distance "a" away before it travels "L+a" to mirror M.

The correct answer simply states this: "C. The object (the light source) is at a distance of L +a in front of Mirror M. Since the mirror is flat, the image will appear to be at this same distance, L + a, behind Mirror M. "

Isn't it neglecting the vertical distance a? Shouldn't D be correct? I like to use extremes to think about it. If light source was actually infinite below from mirror P, which it still only appear L + a behind mirror M? I don't think it would. The distance it is from mirror P should be taken into account for the distance away it appears to be.

 

[NextStepTutor_2]

Its an optics passage, so I'll do must best to make a diagram here.

Mirror M....................................................Partial Mirror P (slanted)
|MM|-------->-------------------------[W]-------<-----\MP\----->-----[Eye]..._
...............................................................................................^..................................|
...............................................................................................^..................................a
..........................................................................................[light]............................. |
.
|<-------------------L------------------->|<-------a------->|<-------a------>|

**The dots are just place fillers. Look at the bolded dashes for the apparatus

So there is Mirror M which is a plane mirror. Mirror P is a partial mirror and slanted toward Mirror M. W is a wheel with slits, which is unimportant for my specific question. The light source comes from below MP, bounces to MM, and travels back through MP to the eye. The vertical distance from the eye to the light source is a, horizontal distance from eye to light is a, and the distance from the wheel to MM is L.

Q4: Seen through the eyepiece, the light source will appear to be located at a distance of:
A. L- 2a in front of Mirror M.
B. L-a behind Mirror M.
C. L + a behind Mirror M.
D. L + 2a behind Mirror M.

Okay, I put D. The answer is completely in the relation to the mirror M, so I started there. If you travel to the light source from Mirror M you go across "L", "a", and then the vertical distance "a" to the light source. Thus, L + 2a behind the mirror because the image MM is seeing from MP is distance "a" away before it travels "L+a" to mirror M.

The correct answer simply states this: "C. The object (the light source) is at a distance of L +a in front of Mirror M. Since the mirror is flat, the image will appear to be at this same distance, L + a, behind Mirror M. "

Isn't it neglecting the vertical distance a? Shouldn't D be correct? I like to use extremes to think about it. If light source was actually infinite below from mirror P, which it still only appear L + a behind mirror M? I don't think it would. The distance it is from mirror P should be taken into account for the distance away it appears to be.

Hi @KoalaT!

For flat mirrors the Law of Plane Mirrors dictates that the image is always the same distance behind the mirror as the object is in front of the mirror.

This link HERE has a pretty good summary of mirrors and MCAT optics.

Hope this helps, good luck!

 

[KoalaT]

Hi @KoalaT!

For flat mirrors the Law of Plane Mirrors dictates that the image is always the same distance behind the mirror as the object is in front of the mirror.

This link HERE has a pretty good summary of mirrors and MCAT optics.

Hope this helps, good luck!

Yes I understand that. But the image was also an addition distance of "a" below the slanted plane mirror. So wouldn't the object mirror M is looking at (the image of mirror P) already appear to be a distance a "a" away before it then travels the other distance of L and a?

 

[NextStepTutor_2]

Yes I understand that. But the image was also an addition distance of "a" below the slanted plane mirror. So wouldn't the object mirror M is looking at (the image of mirror P) already appear to be a distance a "a" away before it then travels the other distance of L and a?

If I understand your diagram correctly, the object of the plane mirror comes from image/reflected light from the slanted mirror, which is L+a distance away from MM. The light that hits MM originates from the reflection in the slanted mirror. Which why that is what the distance through the eyepieces will look like.

Hope this helps, good luck!