Of course you can. The prism is likely to disperse the light - that is, different colors will be refracted by different amounts. That means you don't just get "a" refractive index, but with careful experimental setup you will get the entire refractive index curve: a different value for every wavelength / color.
Setup: a narrow beam of white light incident on the prism. Measure the position where a particular color hits a screen far away. Calculate the angles, use Snell's law to compute $n$. Repeat for every color. If you know the wavelength of the colors (you can look these up) you can create a graph of refractive index vs wavelength; if you don't, you can create a table.
The precision of a setup like this is really only limited by your ability to set it up carefully; and the size. Usually, dispersion is strongest when you are working close to the critical angle of a configuration - this is why a diamond is cut the way that it is. In the diagram above, you are most likely far from the critical angle. This means that the differences in refractive index will matter less.
To get an accurate measure of refractive index, you would like to find the critical angle - that is the angle at which you get total internal reflection. If you know the internal angle at which this happens, you can compute the refractive index directly from Snell's law:
$$\frac{n_1}{n_2}=\frac{\sin\theta_2}{\sin\theta_1}$$
At the critical angle, $\sin\theta_2=1$ and $n_2=1$, so $n_1=\frac{1}{\sin\theta_1}$.
By looking at the color that just gets out of the prism, you find the color of the one that just didn't make it... and that is the color for which you thus find $n$. Now the tricky thing is that in order to do this experiment, the light will also bend at the other interface (the input face of the prism) which makes the math a bit more interesting.
If that is the setup you want to explore and you can't figure out the diagram / equations, please post a comment.
I don't know the "official" answer but here is what I might try. I am hoping that others will contribute to make this a "good" answer.
First - we were not told whether the wavelength of the laser is transmitted at all by the blue foil; but since blue foil typically absorbs red light, and most laser pointers are red (I have a blue one but they are expensive!) I will assume we have no transmission.
That means we need to determine the answer with reflection. The Brewster angle may come to our rescue here. Since a laser beam is polarized, there is a certain angle for which we see no reflection from a surface when the polarization is in the plane containing the normal to the surface and the path of the beam. It should be fairly easy to set up the laser pointer at the Bragg angle (just look at the reflected spot and play around with both the angle of incidence, and the rotation of the laser pointer). Use the ruler to determine the angle (I assume you are allowed a calculator for this exercise - or trig tables, or a good slide rule. This was not specified. If not, then some origami on the graph paper will get you a pretty good goniometer...)
This will give us one data point: since the Brewster angle $\theta_B$ at the interface of materials with refractive index $n_1$ going to $n_2$ is given by
$$\theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right)$$
the refractive index $n_2$ of the slide glass is given by
$$n_g = \tan\theta_B$$
Now comes the tricky part: we want to try to find the point of partial extinction of the second reflection - the one off the interface between the glass and the foil.
For this we want to carefully plot the intensity of the laser beam as a function of angle; I suspect we are looking for a secondary dip in the curve where the reflection from the back surface of the glass / foil interface is completely gone. This will happen when the internal angle ($\theta_2$ in my diagram) obeys the Brewster angle relationship.
We rewrite that relationship as
$$n_{foil} = n_g \tan\theta_2$$
and we know from Snell's Law that
$$\frac{\sin\theta_1}{\sin\theta_2}=\frac{n_2}{n_1}$$
It is possible that there is a range of refractive index values for which there is no solution. Specifically, the angle $\theta_2$ is limited by Snell's Law to be less than $\sin^{-1}\frac{1}{n_g}$, so this approach will not work if
$$\sin\tan^{-1}\frac{n_{foil}}{n_g} \lt \frac{1}{n_g}$$
This answer is not guaranteed to be error free... I need to eat something and revisit this (hope there will be some constructive comments / edits in the meantime).
UPDATE
After reading Chris Mueller's answer, I decided to have another go and plot these curves for the reflected power given two different refractive indices, $n_1$ and $n_2$ (where there is a further $n_0=1$ for air).
Using the Fresnel equations we can compute the sum of reflected intensity for both the "front" and "back" reflections - assuming that the second reflection will in turn be partially reflected at the glass-air interface. There will be multiple internal reflections - each a small fraction of the previous one. There may also be reflections off the back of the foil; but as I said before I am going to assume that red laser pointer light is completely absorbed by the blue foil. The whole system now looks like this:
Then we have to use the following expressions a number of times (suffixes $s$ and $p$ represent polarization - either parallel to the surface, or perpendicular) - I am putting the incident intensity $i_0=1$ for simplicity (note - updated to show coefficients for intensity not amplitude with thanks to Rob Jeffries for pointing out my mistake):
$$R_s=\left|\frac{n_0 \cos\theta_0-n_1\cos\theta_1}{n_0 \cos\theta_0+n_1\cos\theta_1}\right|^2\\
T_s = 1 - R_s$$
$$R_p=\left|\frac{n_0 \cos\theta_1-n_1\cos\theta_0}{n_0 \cos\theta_1+n_2\cos\theta_0}\right|^2\\
T_p = 1 - R_p$$
Now the total reflected intensity is an infinite sum: if we put the coefficient of reflection at the first interface as $R_1$, the second as $R_2$, and the reflection on the inside of the glass-air interface as $R_3$, then we can write for the corresponding transmissions $T_1 = 1 - R_1, T_3 = 1 - R_3$ (we are assuming everything that is transmitted into the foil is absorbed). Then we have for the total reflected intensity:
$$\begin{align}R_t &= R_1 + (T_1 R_2) T_3 + T_1 R_2 R_3 R_2 T_3 + ...\\
&=R_1 + T_1 R_2 T_3 \left(1 + R_2 R_3 + (R_2 R_3)^2 + ...\right)\\
&= R_1 + \frac{T_1 R_2 T_3}{1 - R_2 R_3}\\
\end{align}$$
This gets pretty messy pretty quickly - but that's why we all carry little supercomputers in our pockets these days...
I am going to assume that the laser is polarized and aligned per the above, so we can use just the expression for $R_s$; then something very interesting happens. At the Brewster angle, we will see a minimum in the reflectance - but it will not be zero, since the transmitted beam will undergo partial reflection at the interface between glass and foil. So while $R_1$ in the above expression would be 0, the other terms are not. In fact, it's quite easy to evaluate the expression for a range of refractive index values: and when you do this, you get the following plot:
The Python code I used to generate this plot (note - this does not take account of the correction for the square-of-the-amplitude pointed out above so it will give the wrong quantitative result although it shows the null at the right place):
# use Fresnel to compute reflection from composite surface
import numpy as np
import matplotlib.pyplot as plt
import math
def fresnel(theta1, n1, n2):
n12 = n1 / n2
st = n1/n2 * math.sin(theta1)
if st>1:
return(1, 0)
else:
theta2 = math.asin(st)
r = np.abs( (math.cos(theta1)-math.cos(theta2)*n12)/(math.cos(theta1)+math.cos(theta2)*n12))
return (r, theta2)
theta = np.arange(0, math.pi/2, 0.01)
n2 = np.arange(1, 3, 0.01)
theta0 = math.atan(1.52) # start at Brewster angle
(r1, theta1) = fresnel(theta0, 1.0, 1.52)
t1 = 1 - r1
# second reflection: range of values of n
rt2 = np.array([fresnel(theta1, 1.52, n) for n in n2])
r2 = rt2[:,0]
# probability of light escaping:
(r3, theta3) = fresnel(theta1, 1.52, 1.0)
t3 = 1 - r3
# combine the geometric series:
power = r1 + (t1 * r2 * t3) / ( 1 - r2 * r3)
# plot the result:
plt.figure()
plt.plot(n2, power)
plt.xlabel('foil index')
plt.ylabel('reflected power')
plt.title('compound reflection')
plt.show()
Now if the refractive index of the foil is greater than the refractive index of glass, this gives us an easy way to determine the value by looking at the reflected power: over the range of values I considered, it is almost a straight line.
However, if $n_2 < n_1$, we have that funny curve to the left of the Brewster dip, which is not very helpful. I am still puzzling over how you would deal with that case.
Best Answer
To form the final image the objective of the travelling microscope collects a cone of rays with a range of small angles of incidence (as assumed in the derivation of the formula for refractive index) from a point on the object and not just one ray with an incident angle of zero degrees.
Update as a result of a comment from @SivaManasan
The derivation of the formula for the refractive index is done in many books and in many Internet sites but they all make an approximation of the type if angle $\theta$ is small $\sin \theta \approx \theta,\: \sin \theta \approx \tan \theta $ etc and here is one of those derivations.
My point about the cone of rays I have tried to illustrate below.
The first thing you need to know is that you only move the travelling microscope you make no adjustments to it so the distance between the object and the final image is fixed.
These are distances $CC'$ and $DD'$ in my diagram.
The other important property of the microscope is that its depth of field (distance between the nearest and the furthest objects that are in acceptably sharp focus in an image) is small.
Without the glass block (grey rays) the cone of rays $ACB$ originating from a point object are focussed by the microscope lens $AB$ to form an image in the plane of the cross wires of the microscope at $C'$.
Putting the glass block in the way effectively moves the observed position of the object for $C$ to $D$ and now $D$ acts as the object for the microscope lens.
To focus the cone of rays apparently originating from point $D$ the microscope lens has to be moved to position $A'B'$ which is distance $h$ in my diagram.
On moving the microscope (lens) the cone of rays $A'DC'$ are now focused by the lens at $D'$.
Note that all the rays (except the zero incidence ray) undergo refraction (bending of a light ray as it travels from one medium to another) and so the derived formula for the refractive index is valid particularly as the angles involved are small because of the small diameter of the microscope lens.
There are other methods with pins etc but one of the main reasons why Indian sites have a lot of reference to to this method of measuring refractive index might be that the topic is in the syllabuses of Indian examination boards whereas, for example in the UK, this method has been omitted in favour of other methods which mostly rely on the measurement of angles.