TPR Mirror Physics Question

[Dazzled03]

---

Members don't see this ad.

From Passage 55, Question 4 in the Hyperlearning Science workbook:

I attached a picture of the figure that the question is referring to.

Q: "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

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

Just wanted to see if someone can further break down the reasoning behind this for me? Does the distance (a) from the light source to mirror P cancel out with the distance (a) the light travels from mirror P to the eye-piece?

I understand that for a plane mirror, the object distance is equal to the image distance. But I don't see why the object distance wouldn't include the distance the light travels from the source to mirror P?

 

[iliveinmyguitar]

I know this is old but can we please break this one down. I thought the right answer was D because I felt like we should include the distance from the light source to mirror P, from mirror P to the wheel, and from the wheel to mirror M. No?

 

[amy_k]

For the image distance we only need to account for the distance from the partial mirror to mirror M, which is L + a. The partial mirror splits the light source beam equally and this split beam is the "new" incident ray, traveling a distance of L + a to the mirror M.

 

[iliveinmyguitar]

For the image distance we only need to account for the distance from the partial mirror to mirror M, which is L + a. The partial mirror splits the light source beam equally and this split beam is the "new" incident ray, traveling a distance of L + a to the mirror M.

coool! thanks alot