I think all you need to solve this is a diagram, and unlike me, you need to remember how big an angstrom is. Right: wikipedia says its 10^-10 m So 6000 angstrom is 6*10^-7 m When the light is virtually parallel it is in phase, so there is a bright fringe in the center. If the width of a fringe is 0.1 degrees, it means the next bright fringe is 0.1 degrees further up. At that point the light is in phase again. If you draw this diagram and note that this is the first time it is in phase again you will see that there is a right triangle: (I, however, am just imagining it). The hypotenuse is the distance between the slits. The very short base is one wavelength of the light travelling from the lower slit upwards at a shallow 0.1 degree angle. (upwards on the diagram this is). The third side of the triangle runs from the upper slit to the light beam from the lower slit where it meets at a right angle. The top angle in this triangle is 0.1 degrees. You can see that by imagining lifting the light beam up and down and seing that the third side rotates the same amount. So SOHCAHTOA says that 6 * 10^-7 divided by spacing = sin 0.1 degrees. spacing = 6 * 10^-7 meters / sin 0.1 degrees = 3.4 mm That seems a little too small to be practical to me, so you better check the math with a real diagram.

use the formula theta=(1.22xlambda)/d to find the interslit distance d=(1.22x6000)/0.1; d=73200 angstrom

first convert the 0.1 degrees into radians (0.1*3.14)/180 which is equal to omega i.e. angular width which is further equal to (wavelength)/slit distance.... so for slit distance the formula becomes (lamda)/(omega) the answer to this problem is 3.4*10^-4 m