Our instructor told us about this phenomenon:

enter image description here

The back of the glass slab is silvered, and M' is the front-shifted image of the silvered surface. While explaining reflection off of this, he stated the object distance to be; $$u = x + t/n$$ Apparently, light reflects off the image of the silvered surface, forming an image equally spaced behind M'.

How is this possible? I have heard of virtual objects, but can they serve as mirrors?

  • $\begingroup$ Refer shift due to glass slab/Apparent depth. Anyway, I don't know why we considered shift of mirror, the incident rays of object is the one that undergoes shift. So the object distance would be (x+t)-(t/n) from the position of the old mirror, it would reflect back and the image would undergo a shift of t/n again in the direction of incident rays. $\endgroup$
    – user600016
    Apr 27, 2020 at 7:47
  • $\begingroup$ @user600016: Please answer using the answer box below and not in the comments. Comments which tend to answer the question might be deleted. $\endgroup$
    – Vishnu
    Apr 27, 2020 at 7:50
  • $\begingroup$ I have no problem in doing so @guru vishnu! But there are a few pedantics who always insist on explaining /expanding each line, and I currently don't have the time to do so. $\endgroup$
    – user600016
    Apr 27, 2020 at 9:00

1 Answer 1


As you probably might know that the shift caused by a slab of refractive index $n$ of thickness $t$ is:

$$s=t\left(1-\frac 1 n\right)$$

What this means is, the effect of the slab is same as shifting the object towards the mirror by an amount of $s$ and then removing the slab. Or in other words, shifting either the mirror or the object by $s$ after removing the slab has the same effect of the presence of the slab.

Thus, the shift of "virtual objects" due to the presence of slab is equally applicable to the shift of mirrors.

  • $\begingroup$ That's fine, but in the derivation of image distance following this, M' is actually treated as a real mirror; as if the reflection took place from it. How can a reflection of a mirror actually serve as one $\endgroup$
    – harry
    Apr 28, 2020 at 3:10

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