Given a square $ABCD,$ with $AB=1$ mark the midpoints $M$ and $N$ of $AB$ and $BC,$ respectively. A lasar beam shot from $M$ to $N,$ and the beam reflects of $BC,CD,DA,$ and comes back to $M.$ This path encloses a smaller area inside square $ABCD.$ Find this area. 
1. **Identify the midpoints and initial setup:**
   - Given a square \(ABCD\) with side length \(AB = 1\).
   - Mark the midpoints \(M\) and \(N\) of \(AB\) and \(BC\), respectively.
   - Coordinates of points:
     - \(A = (0, 0)\)
     - \(B = (1, 0)\)
     - \(C = (1, 1)\)
     - \(D = (0, 1)\)
     - \(M = \left(\frac{1}{2}, 0\right)\)
     - \(N = \left(1, \frac{1}{2}\right)\)

2. **Path of the laser beam:**
   - The laser beam starts at \(M\) and travels to \(N\).
   - It reflects off \(BC\) and travels to a point on \(CD\).
   - It reflects off \(CD\) and travels to a point on \(DA\).
   - It reflects off \(DA\) and returns to \(M\).

3. **Determine the path and enclosed area:**
   - The path of the laser beam forms a smaller square inside the original square \(ABCD\).
   - The vertices of this smaller square are the midpoints of the sides of \(ABCD\):
     - \(M = \left(\frac{1}{2}, 0\right)\)
     - \(N = \left(1, \frac{1}{2}\right)\)
     - Midpoint of \(CD\) is \(\left(\frac{1}{2}, 1\right)\)
     - Midpoint of \(DA\) is \(\left(0, \frac{1}{2}\right)\)

4. **Calculate the side length of the smaller square:**
   - The distance between any two adjacent midpoints is the side length of the smaller square.
   - For example, the distance between \(M\) and \(N\):
     \[
     \text{Distance} = \sqrt{\left(1 - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - 0\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}
     \]

5. **Calculate the area of the smaller square:**
   - The area of a square is given by the side length squared.
   - Side length of the smaller square is \(\frac{\sqrt{2}}{2}\).
   - Therefore, the area of the smaller square is:
     \[
     \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2}
     \]

The final answer is \(\boxed{\frac{1}{2}}\).