Answer
a. $a=4.17\times10^{-4}m $.
b. $a=4.17\times10^{-2}m $.
c. $a=4.17\times10^{-7}m $.
Work Step by Step
Use equation 36.3 because the angle is small.
$$y_1=\frac{x \lambda}{a}$$
In each case, the width of the central maximum, 6.00 mm, is $2y_1$. Therefore, $y_1=3.00\times10^{-3}m.$
a. $$a=\frac{x \lambda}{y_1}=\frac{(2.5m) (500\times10^{-9}m)}{ 3.00\times10^{-3}m }=4.17\times10^{-4}m $$
b. $$a=\frac{x \lambda}{y_1}=\frac{(2.5m) (50\times10^{-6}m)}{ 3.00\times10^{-3}m }=4.17\times10^{-2}m $$
c. $$a=\frac{x \lambda}{y_1}=\frac{(2.5m) (0.5\times10^{-9}m)}{ 3.00\times10^{-3}m }=4.17\times10^{-7}m $$
