Question

A 0.45-mm-wide slit is illuminated by light of wavelength 590 nm. What is the width (in mm) of the central maximum on a screen 1.5 m behind the slit?

Answer 1

Answer:

$w = \frac{2(590x10^{-9}) (1.5m)}{0.45x10^{-3}}=3.93 x10^{-3} m$

And if we convert this into mm we got:

$w= 3.93 x10^{-3} m *\frac{1000mm}{1 m}= 3.93 mm$

Step-by-step explanation:

For this case we have the following info givenL

$\lambda = 590 nm = 590 x10^{-9}m$ represent the wavelength

$s = 0.45 mm *\frac{1m}{1000 mm}= 0.45x10^{-3} m$ represent the width of the single slit

$L = 1.5 m$ represent the longitude

And we want to find the maximum width of the central maximum in mm, so we can use the following formula:

$w s = 2 \lambda L$

And if we solve for w the width we got:

$w = \frac{2\lambda L}{s}$

And replacing we got:

$w = \frac{2(590x10^{-9}) (1.5m)}{0.45x10^{-3}}=3.93 x10^{-3} m$

And if we convert this into mm we got:

$w= 3.93 x10^{-3} m *\frac{1000mm}{1 m}= 3.93 mm$