ASCII Doodling: Laser in a Box

Pyth, 43 41 39 bytes

K*\#+2QKVvzp<*QXX*dyivzQN\\_hN\/Q\#\#)K

Try it online: Pyth Compiler/Executor. Input the numbers in the following order: height first line, width second line.

Thanks to isaacg, who helped saving two bytes.

Explanation:

My solution doesn't trace the laser, it uses a simple pattern that includes the gcd. If m, n are the dimensions of the box, let d = gcd(m, n). The size of the pattern is exactly 2*d x 2*d.

E.g. the repeating pattern for 7 5

#########
#\/\/\/\#
#/\/\/\/#
#\/\/\/\#
#/\/\/\/#
#\/\/\/\#
#########

is

\/
/\

(gcd(7, 5) = 1, size of pattern is 2 x 2)

And the repeating pattern for 22 6

########################
#\  /\  /\  /\  /\  /\ #
# \/  \/  \/  \/  \/  \#
# /\  /\  /\  /\  /\  /#
#/  \/  \/  \/  \/  \/ #
#\  /\  /\  /\  /\  /\ #
# \/  \/  \/  \/  \/  \#
########################

is

\  /
 \/ 
 /\
/  \

(gcd(22, 6) = 2, size of pattern is 4 x 4)

My solution does the following thing for each of the lines: it simply generates one line of the pattern, repeats it a few times and cut's it at the end so that it fit's into the box.

K*\#+2QK   implicit: Q is the second input number (=width)
K          K = 
 *\#+2Q        "#" * (2 + Q)
       K   print K (first line)

Vvzp<*QXX*dyivzQN\\_hN\/Q\#\#)K  implicit: vz is the first input number (=height)
VQ                               for N in [0, 1, ..., vz-1]:
           ivzQ                             gcd(vz,Q)
          y                               2*gcd(vz,Q)
        *d                           string with 2*gcd(vz,Q) space chars
       X       N\\                   replace the Nth char with \
      X           _hN\/              replace the -(N+1)th char with /
    *Q                               repeat Q times
   <                   Q           only use the first Q chars
  p                     \#\#       print "#" + ... + "#"
                            )    end for
                             K   print K

C, 256 bytes

f(w,h){int i,j,x=1,y=1,v=1,u=1;char b[h+2][w+3];for(i=0;i<w+3;i++)for(j=0;j<h+2;j++)b[j][i]=!i||!j||i>w||j>h?i>w+1?0:35:32;while((x||y)&&(x<=w||y<=h))v=x&&w+1-x?v:(x-=v,-v),u=y&&h+1-y?u:(y-=u,-u),b[y][x]=v/u<0?47:92,x+=v,y+=u;for(i=0;i<h+2;i++)puts(b[i]);}

I can probably get this under 200, and I'll add an explanation later, but I might have a paper due in a few hours I should be doing instead.

J, 85 bytes

Let g = gcd(w,h). The function fills the elements of a w/g by h/g matrix with g by g tiles, having /'s and \'s in their diagonal and anti-diagonal. The resulting 4D array is raveled into a 2D one (the inside of the box) then surrounded with #'s. (The numbers 0 1 2 3 are used instead of [space] / \ # and the numbers are changed to characters at the end.)

A direct position based computation of inside coordinate could maybe yield a bit shorter solution.

' \/#'echo@:{~3,.~3,.3,~3,,$[:,[:,"%.0 2 1 3|:((,:2*|.)@=@i.@+.){~[:(2&|@+/&:i.)/,%+.

Usage:

   6 (' \/#'echo@:{~3,.~3,.3,~3,,$[:,[:,"%.0 2 1 3|:((,:2*|.)@=@i.@+.){~[:(2&|@+/&:i.)/,%+.) 22
########################
#\  /\  /\  /\  /\  /\ #
# \/  \/  \/  \/  \/  \#
# /\  /\  /\  /\  /\  /#
#/  \/  \/  \/  \/  \/ #
#\  /\  /\  /\  /\  /\ #
# \/  \/  \/  \/  \/  \#
########################

Try it online here.