Question

Young's double slit experiment is carried out using microwaves of wavelength λ=3 cm. Distance between the slits is d=5 cm and the distance between the plane of slits and the screen is D=100 cm. Then what is the number of maximas and their positions on the screen?

Answer 1

Hint: In Young’s double-slit experiment interference of light takes place
And due to constructive and destructive interference of light, we get the pattern on the screen.

Complete step by step answer:Let the density of the planet be \[\rho \]and let the density of Given values, λ=3 cm, d=5 cm, D=100 cm.
To find out the number of maxima we know,
maximum path difference = d=5 cm.
Thus, in this case we can have only three maximas, one central maxima and two first maxima.

b)- Now to find maximum intensity, , we know the path difference between the two interfering waves is given by the formula, \[{{S}_{2}}p-{{S}_{1}}p=\lambda \]where the LHS represents the path difference between the two interfering waves. Now in terms of y, d and D we can write it as:
\[\sqrt{{{(y+\dfrac{d}{2})}^{2}}+{{D}^{2}}}-\sqrt{{{(y-\dfrac{d}{2})}^{2}}+{{D}^{2}}}=\lambda \]
Substituting d=5cm, D = 100cm and \[\lambda \]= 3cm we get, \[\lambda =\pm 75cm\]
Thus, the three maximas will be at:
y=0
y=75cm and y=-75 cm
all distances are measured in cm.

Additional information- When an interference pattern is observed on the screen one distinct feature is all the bright bands are equally bright and all the dark bands are equally dark.

Note- Interference of light is a phenomenon in which two or more waves interfere with each other and there is redistribution of energy. This is one of the most amazing phenomena to be ever observed and the pattern is worth watching.