Question on mirror and lens combinations

[BlueCerealBox]

Homework Statement

The question is attached in the jpeg file.

Homework Equations

1/do + 1/di = 1/f

The Attempt at a Solution

First reflection : 1/100 + 1/di = 1/80 , di = 400cm ( To the right of the lens )

This means that the image created is a virtual image since it is behind the lens.

So for second reflection : -1/300 + 1/di = -1/50 , di = -60cm ( To the right of the mirror )

The image is virtual , therefore for last reflection : -1/160 + 1/di = 1/80 , di = 53.3cm

I can't seem to be able to get the answer , if I took the object distance for the last reflection as positive. I do get the answer. But it does not make sense to me.

I usually have problems with these kind of questions as I'm unsure where is infront or behind the mirror? Is this defined by the original real object that causes the reflections to occur?

 

[Aceix]

Homework Statement

The question is attached in the jpeg file.

Homework Equations

1/do + 1/di = 1/f

The Attempt at a Solution

First reflection : 1/100 + 1/di = 1/80 , di = 400cm ( To the right of the lens )

This means that the image created is a virtual image since it is behind the lens.

So for second reflection : -1/300 + 1/di = -1/50 , di = -60cm ( To the right of the mirror )

The image is virtual , therefore for last reflection : -1/160 + 1/di = 1/80 , di = 53.3cm

I can't seem to be able to get the answer , if I took the object distance for the last reflection as positive. I do get the answer. But it does not make sense to me.

I usually have problems with these kind of questions as I'm unsure where is infront or behind the mirror? Is this defined by the original real object that causes the reflections to occur?

Lenses refract light. And for this convex lens, the image formed will be real because the object distance is greater than the focal length.
Welcome to PF.

 

[hectic_night]

so i use a different sign convention than you , i take the center of the lens as origin and imagine a graph ,so at last reflection the object(the virtual image created by the mirror ) is to the right of the mirror so , i take the focal length of the convex lens as negative (if the center is origin the focal length is towards left so negative )
and the object distance is positive since it is to the right
by using the formula of lens 1/f = 1di - 1/do
i get 1/-80 = 1/v - 1/160
1/-80 + 1/160 = 1/v
after solving u get v = -160
so remember the sign conventions since the center is origin i got a negative value so the image lies to the left of the lens