We can use the same expression that is commonly used to estimate the apparent magnitude of a planet or asteroid in the Solar System*:
$$\boxed{m=5 \log \frac{1329}{d \cdot \sqrt p}+5 \log (D_s \cdot D_e)-2.5\log f(F)}$$
Where:
m is the apparent magnitude
log is the decimal logarithm
d is the diameter in km
p is the albedo
Ds is the distance to the Sun in Astronomical Units (AU)
De is the distance to Earth in Astronomical Units (AU)
f(F) is the phase function
If we want to apply this expression to the James Webb Space Telescope, the values to use are as follows.
As the surface area of the JWST solar shield is $S=21\cdot 14=294$ square meters:
$$d=\sqrt{\frac{4\cdot S}{\pi}}$$
Equivalent diameter "d"
d = 0.01935 km
The satellite hosting the Webb Telescope will be placed in a halo orbit around Lagrange L2 Sun-Earth, a point located 1.5 million kilometers from the Earth (see drawing). The radius of the halo orbit is 0.8 million km.
$$D_e=\sqrt{1.5^2+0.8^2}=1.7$$
De = 0.01136 AU
Ds = 1.01004 AU
As a phase function we will use the usual one:
$$f(F)=\dfrac{1+\cos F}2$$
The phase angle "F" is calculated using basic trigonometry from the drawing
It is obtained:
F = 27.8º
And we see that the contribution of the corresponding summand is almost negligible:
$$-2.5 \log f(F)=+0.06$$
The albedo "p" of the heat shield will be high, to reflect the maximum amount of heat possible. 371
But since the albedo of the solar shield is unknown to me, I add the table below showing the estimated apparent magnitude value, for different albedo values.
The table tells us that the apparent magnitude of the JWST when it reaches its halo orbit around L2 will be of the order of +15 and therefore this should be visible with some powerful telescopes of amateur astronomers.
*Reference: Binary asteroid population. 1. Angular momentum content, by <PERSON> and <PERSON>
Best regards. 758