A particle in a box is measured to not be on the right hand side of the box. How does the wave function collapse? Starting with initial wave function of a particle in a box with width $2\pi$.

We look for the particle in the right half of the box but do not find it there. Does the wave function collapse to the orange or blue wave? Or something else entirely?

 A: What actually determines the form of the collapsed wave function is the part of the Hamiltonian describing the interaction between the particle and the measurement device. Suppose our initial Hamiltonian is
$$
H_0=\frac{p^2}{2m}+V(x)
$$
where
$$
V(x)=\begin{cases}0, 0<x<a,\\ +\infty \text{ otherwise}\end{cases}
$$
Now measure the position of the particle by a measurement device that localizes it in the region $[0,a/2]$, i.e., by suddenly turning on potential
$$
U(x)=\begin{cases}0, 0<x<\frac{a}{2},\\ +\infty \text{ otherwise}\end{cases}
$$
The initial wave function in potential $U(x)$ need now be projected onto the eigenstates of the potential $U(x)$ and normalized, giving us the desired answer. Note however, this is a general but difficult way to get the answer given in the comments by @SolubleFish:

It converges to 2 times the orange one.

Remark:
The total measurement can be thus described by Hamiltonian
$$
H(t)= H_0+U(t)\theta(t),
$$
where $\theta(t)$ is the heaviside step function.
The non-unitarity of the wave function collapse results from the singular nature of the potential $U(x)\theta(t)$ - both in space and time. If instead the potential is finite in the whole region $[0,a]$ and/or if the potential in the part $[a/2,a]$ grows smoothly in time, the evolution would be unitary and the wave function would flow smoothly into the left part of the well.
A: @RogerVadim 's answer uses a (somewhat) realistic model for the measurement process. The Copenhagen interpretation gives some mathematical prescription for measurement, which are simpler and less realistic. Here is what it prescribes :
(Edit : For simplicity, I consider the box centered around zero so $x \in [-\pi/2 ,\pi/2]$)
"Finding the particle on the right" is measured by the observable $\theta(\hat x)$ (with $\theta$ the Heaviside step function). It is a hermitian projection operator, so under the Copenhagen interpretation, if we find it on the right, the wave function collapses to :
$$\psi_R(x) = \frac{1}{\int_0^{\pi/2}|\psi(x)|^2 \text dx}\theta(x)\psi(x)$$
If we don't find it (ie it is on the left), we use the projection operator $1-\theta(\hat x) = \theta(-\hat x)$ and get :
$$\psi_L(x) = \frac{1}{\int_{-\pi/2}^0 |\psi(x)|^2 \text dx}\theta(-x)\psi(x)$$
Since we are starting with an even wave function, the normalization comes out to $2$, so the collapsed wave function is :
$$\psi_L(x) = 2 \theta(-x)\psi(x)$$
ie $2$ times the orange wave-function.
