Question

Light with wavelength 575 nm goes through a single slit of width 0.036 mm and displays a diffraction pattern on a screen 4.2 m away. 1.What is the width of the central bright fringe?
2. What is the angle to the 3rd dark fringe?
3. What is the momentum of a single photon of light with a wavelength of 575 nm?

Answer 1

Answer:

momentum = 1.147 × $10^{-34}$ kg-m/s

Explanation:

given data

wavelength = 575 nm = 575 × $10^{-9}$ m

slit width S = 0.036 mm = 0.036 × $10^{-3}$ m

diffraction pattern D = 4.2 m

to find out

width of central bright fringe, angle of 3rd dark fringe and momentum of photon

solution

we will apply here formula for central bright fringe width formula that is

width = $\frac{2*wavelength*D}{S}$   ...............1

put here all these value

width = $\frac{2*575*10^{-9}*4.2}{0.036*10^{-3}}$

width = 0.134 m

so width of central bright fringe is 0.134 m

and

for angle we will formula

slit width × sin∅ = n × wavelength    ............2

here n is no of dark fringe

put here value and find angle

0.036 × $10^{-3}$ × sin∅ = 3 × 575 × $10^{-9}$

solve and we get

∅ = 2.75°

so angle to the 3rd dark fringe is 2.75°

and

momentum formula is

momentum = $\frac{planck constant}{wavelength}$

here we know planck constant for photon is 6.6 × $10^{-34}$

so

momentum = $\frac{6.6*10^{-34}}{575*10^{-9}}$

momentum = 1.147 × $10^{-34}$ kg-m/s