Question

a single slit diffraction pasttern has the central maximum width of 1 cm. Given the wavelength of 500 nm and slit width of 0.01 mm, what is the distance from the slit to the observation screen? 1 m
0.1 m
71 cm
0.71 m
none of the above?

Answer 1

Answer:

distance between slit and screen, d = 0.1 m

Given:

slit width, x = $0.01 mm = 0.01\times 10^{-3} m$

width of central maximum, t = 1 cm

wavelength, $\lambda = 500nm = 500\times 10^{-9} m$

Explanation:

The first minima will be at the position t:

t = $\frac{1}{2} = \frac{1}{2} = 0.5 cm$

Now, using the formula:

xsin$\theta = n\lambda$

where

n = 1

xsin$\theta = \lambda$

sin$\theta = \frac{\lambda}{x}$

sin$\theta = \frac{500\times 10^{-9}}{ 0.01\times 10^{-3}} = 0.05$

Now, calculation of distance between slit and screen, d:

$d = \frac{t}{sin\theta}$

$d = \frac{0.5\times 10^{-2}}{sin0.05} = 0.1 m$