Help calculating the power function at a point $\mu=0.5$ for the normal distribution For some reason, I've been stuck on this question, even though other questions of this type weren't that much of a problem to me before (with for instance the binomial distribution). The situation is as follows:
Suppose $X_1...X_{25}$ is a sample from an $N(\mu,2^2)$-distribution. We would like to test for $H_0: \mu\leq 0$; $H_1:\mu>0$, with significance $a_o=0.05$.
Calculate the power function of this test at the point $\mu=0.5$.
My approach:
Earlier in this problem I had to calculate the critical values and found the following: $K_T = \{T\in \mathbb{R}:T\geq 1.64\}$} and $K=\{\mu\in\mathbb{R}:\mu\geq0.656\}$} where $T=\sqrt{n}\frac{\mu-\mu_0}{\sigma}$
For $H_0$ we use $\mu_0=0$ and find (with a table or calculator) these values for $c$ in $P(T\geq c)\leq0.05$.
Now we have to use $\mu_0=0.5$. Given this information, we have to calculate $P_{0.5}(X\in K)$ which is the power function. I thought of two things, but they both give me wrong answers. 
First I tried to just calculate $P(T\in K_T$) given $\mu=0.5$, but I stumbled upon a problem; I didn't seem to have used the fact that $\mu=0$ in my calculation  for the critical values for $K_T$. I just chose $T=\sqrt{n}\frac{\mu-\mu_0}{\sigma} =\frac{5}{2}\mu$ and said that $T\sim N(0,1)$ and proceeded to use the table. Only for $K$ I used that $T=\frac{5}{2}\mu\geq1.64 \iff \mu\geq0.656$. (so here we used $\mu_0=0\implies T=\frac{5}{2}\mu$)
So, changing to $\mu_o=0.5$ didn't really give me anything in this calculation, if I would just put it in there I would get that $P(T\in K_T)=0.05=a_0$ again.
Then I tried to calculate 
$$P(X\in K)=P(N(0.5,2^2)\geq1.64)=$$
$$P(4N(0,1)+0.5\geq1.64)=P(N(0,1)\geq 0.285)=0.39$$
This comes closer, but the correction model says it's $0.34$. I did find, that $P(N(0,1)\geq a)=0.34$ for $a=0.41$ which is $1.64/4$, the original critical value divided by $4$.
I really feel like I'm missing a (relatively simple) key element of the way to approach this problem. Any corrections, tips, hints or explanations are highly appreciated!
 A: Comment: The power computation in the Answer by @a_statistician
is correct (+1) for $n=25, \alpha= .05, \sigma = 2,$ and $\Delta = \mu_a - \mu_0 = 0.5$ (one-sided alternative). 
Because the power for these parameters seems lower than is usually considered desirable, I
show power functions from Minitab software that illustrate what is required for
larger power values. I use both $n = 25$ (lower curve) and $n = 150$ (upper).
Power and Sample Size 

1-Sample Z Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.05  Assumed standard deviation = 2


            Sample
Difference    Size     Power
       0.5      25  0.346475
       0.5     150  0.921760


From the upper curve we see that the power increases from about 0.35 to about 0.92
if we increase $n$ from 25 to 150. Alternatively, from the lower curve we see
that with $n = 25$ the power against the alternative $\mu_a = 1.2$ is about 0.9.
Notes: (a) The slight difference between power 0.3465 and 0.3483 seems to be because of rounding necessary to use a printed standard normal table.
(b) To be fussy, the vertical axis might be labeled "P(Reject)" instead of
"Power": Except for one point, the points on the curve are power, that is, $P(\text{Reject } H_0 | \mu_a).$ The exception is that the left-most point
shows the significance level, $\alpha = P(\text{Reject } H_0 | \mu_0 = 0).$ 
(c) Many statistical
software programs include procedures for making power curves. R has a library
devoted to power curves.
(d) Some textbooks give a formula for the power of such a (one-sided) z-test:
$$ \pi(\mu_a) = P\left(Z > z_{\alpha} - \frac{|\mu_0-\mu_a|}{\sigma/\sqrt{n}}\right).$$
The Answer by @a_statistician can be used as a template for a proof of this formula. With a slight change in notation, I
copied it from Ott & Longnecker, 5e, Chapter 5. It is the basis for making power curves such as the one above. In R, the specific computation in your Question is as follows:
1 - pnorm(qnorm(.95) - .5/(2/5))
[1] 0.3464755

A: Under $H_o$, find the critical value (C) for $\bar X$, which is
$$\Pr(\bar X > C) =\Pr(\frac {\bar X}{2/5}>\frac C{2/5}) = 0.05$$
We have $\frac C{2/5} = 1.64$, so $C= 1.64\times \frac 25=0.656$. You did it correctly.
Then think about if $H_a: \mu = 0.5$ is true, $\bar X \sim N\left(0.5, \frac {2^2}{25}\right)$. 
The Probability of $\bar X>0.656$, which is power. is
$$\Pr(\bar X >0.656) = \Pr(\frac{\bar X -0.5}{2/5}>\frac {0.656-0.5}{2/5})$$ 
It equals $Z = (0.656 - 0.5) \times 5/2 = 0.39$
From standard normal distribution, $\Pr(Z>0.39) = 0.3483$
