I think the distance that should be used is the distance that one expects to be from the game you are hunting. Before taking a shotgun for a gobbler or even for ducks or other animals, you need to see how your gun performs by patterning it at various ranges with the load you want to use.
The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.
Answer:
Explanation:
The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under
Total time = Time taken through ice + Time taken through quartz
Time taken through ice = Thickness of ice / (speed of light in ice)
Thus in the same time the it would had covered a distance of
![Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3DTotaltime%5Ctimes%20V_%7Bvaccum%7D%5C%5C%5C%5CDistance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Cmu%20_%7Bice%2B1.50%5Cmu%20_%7Bquartz%7D%7D%5D)
we have
Applying values we have
![Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Ctimes%201.309%2B1.50%5Ctimes%201.542%5D)
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
