# See the opposite wall on a mirror [closed]

I'm in tenth standard. This is a higher-order-thinking-skills Q I found in a book. One is supposed to use laws of reflection ($\angle i = \angle r$). You can also use mathematical concepts like similarity and trigonometry.

You are in the centre of a room (I assume cuboidal). There is a plane mirror hanging on one wall. If your eyes are $h$ distance from the ground, what is the minimum height of the plane mirror required to see the entire wall behind you.

Initially, I didn't get the answer. The answer in the book says $h/2$.

Then I came up with a solution that said:

$$Mirror\ height=\frac{Wall\ height}3$$

and I thought that the mirror height was independent of $h$.

I just had a thought. Maybe we are allowed to turn the mirror to an angle. But that is not clearly specified and I don't know how it would benefit us.

• Try drawing a diagram, you should then 'see the light', no pun intended.
– Gert
Oct 1, 2015 at 15:57
• The answer obviously isn't $h/2$ because your eyes could be a distance $h \approx 0$ from the ground if you were lying down and the mirror height in that scenario can't also be zero or you wouldn't be able to see the top of the rear wall. Actually I get the same as you - wall height/3. Oct 1, 2015 at 16:12
• @JohnRennie Okay, thanks. Are such questions not allowed on this site? Because the reason given for closing is a homework question. Oct 2, 2015 at 6:23
• @JohnRennie a) I did put some effort in finding the answer b) The question is useful to future users. c) It is not exactly a homework question, it's just a question from a book. Even my teacher was not entirely convinced that the answer in the book was wrong. Oct 2, 2015 at 6:25
• @ghosts_in_the_code: It's hard to see how the answer would be of use to anyone unless they are either (a) using the same book as you or (b) encounter the same question in an exam. This is going to be a tiny subset of site users, and thats why we disapprove of such answers. If the question explored important concepts in light propagation that would be a different matter. Still, I sympathise and that's why I provided a brief answer in a comment. Oct 2, 2015 at 6:31

To see complete wall (of height $$l$$) behind himself, a person requires a plane mirror of at least $$\frac{1}{3}$$ of the height of wall. It should be noted, as mentioned in the problem, that person is standing in the middle of the room.

The ray diagram is more or less like this.

The next diagram is however not that good . But for explanation consider it to work good.

From picture, we proceed to prove as follows:

Say $$MM' = r \,\ unit \,\ and \,\ EA = EG = HF = BD = MM' = r .....(i) \,\ (\,\ by \,\ construction)$$

In $$\Delta$$ RMB, C is mid-pt. of RB and CA is parallel to RM.

So A is mid-pt. of BM.

Similarly, I is mid-pt. of QH implies that G is mid-pt. of HM'.

Again in $$\Delta$$ MFB, A is mid-pt. of BM and AE is parallel to BF.

So, E is mid-pt. of MF and from the mid-pt. theorem on $$\Delta$$ M'DH, E is mid-pt. of M`D.

Therefore, $$2EA = FB \,\ and \,\ 2GE = HD \,\ implies \,\ HD = FB = 2r ....(ii)$$

But from (i) and (ii), we can say that $$HD = HF + FD = r + FD = 2r$$ implies $$FD = r$$

Therefore $$HF = FD = BD = r \,\ i.e. \,\ HB = HF + FD + BD = 3r$$

Therefore $$l = 3r$$ implies $$r = \frac{l}{3}$$

EDIT: Now according to the question, the person's eyes are at a height $$h$$ from ground.

So, $$EA = r \,\ and \,\ AC = h-r$$

And from $$\Delta$$ BRM, $$MR = 2AC = 2(h-r)$$ Similarly M'Q = 2(h-r)

Now as assumed HB = QR = l, $$MR + MM' + M'Q = QR = l \,\ implies \,\ 2(h-r) + r + 2(h-r) = l$$ or, $$4h - 3r = l$$

And as earlier proved $$r = \frac{l}{3}$$ we have $$4h - 3r = 3r \,\ implies \,\ 6r = 4h \,\ or, \,\ r = \frac{2h}{3}$$ So your answer is $$\frac{2}{3}$$rd of the height at which the eye is stationed.

Hope it helps you.

• This is correct (though the derivation could be a lot briefer) but note that in the question $h$ is not the height of the wall. It's the height of the person's eyes above the ground. Oct 2, 2015 at 6:33
• @JohnRennie Oops..missed it. Thanks for pointing it out. I have updated my answer, this time considering $l$ as the height of wall. And ya..the answer could have been briefer..I wrote in too much detail. Oct 2, 2015 at 12:42

The answer is $\dfrac{L}{2}$, where $L$ is the length and height of the wall. When you are standing in such a way that your eyes are at height $h$, the height of the wall above you is $L-h$. Now to see that part, you need a mirror of height $\dfrac{L-h}{2}$ because of $\textit{i=r}$, you will be able to see rest $\dfrac{L-h}{2}$ . By the same logic, the mirror needs to have $\dfrac{h}{2}$ below the level of eyes. Thus, height of mirror must be $\dfrac{L-h}{2} + \dfrac{h}{2} = \dfrac{L}{2}$. Same logic applies to the other dimension of the mirror. So, if it is a $AXB$ wall , then you need $\dfrac{A}{2} X \dfrac{B}{2}$ mirror to watch the wall at your back.

• This is wrong. See Aniket's answer. Oct 2, 2015 at 6:28
• @John Rennie Okay, agree! I realize what mistake I made. Thank you. Oct 2, 2015 at 12:00