Question

If the first minima in Young’s double-slit experiment occur directly in front of the slits (distance between slit and screen $D = 12cm$ and distance between slits $d = 5cm$, then the wavelength of the radiation used can be
A. $2cm$
B. $4cm$
C. $\dfrac{2}{3}cm$
D. $\dfrac{4}{3}cm$

Answer 1

Superposition principle says when two or more waves overlap in space, the resultant disturbance is the sum of the individual disturbances.
In this question, it is given that a first minimum obtained on the screen is directly in front of the slits; hence we have to find the wavelength of the radiation given as $\vartriangle x = \left( {2n - 1} \right)\dfrac{\lambda }{2}$. First we need to evaluate the distance travelled by the radiated light in between the two slits one by one and then, equating their difference with the formula $\vartriangle x = \left( {2n - 1} \right)\dfrac{\lambda }{2}$ to get the result.

Complete step by step solution:
Distance between screen and slit $D = 12cm$
Distance between slits $d = 5cm$
First minima fall directly in front of the slit ${S_1}$ at point P

Distance travelled by wave from ${S_1}$to {S_1}P$$$$ = 12cm
The path travelled by light from slit ${S_2}$to {S_2}P$$$$ = \sqrt {{D^2} + {d^2}} = \sqrt {{{12}^2} + {5^2}} = \sqrt {144 + 25} = \sqrt {169} = 13cm
So the path difference between two minima $\vartriangle x = {S_2}P - {S_1}P = 13 - 12 = 1cm$
Given the point for first minima$\left( {n = 1} \right)$, $\vartriangle x = \left( {2n - 1} \right)\dfrac{\lambda }{2} = \left( {2 \times x - 1} \right)\dfrac{\lambda }{2} = \dfrac{\lambda }{2}$
Hence we can write:

$$\vartriangle x = \dfrac{\lambda }{2} \\\ \dfrac{\lambda }{2} = 1 \\\ \lambda = 2cm \\\$$

Hence the wavelength of the radiation is 2cm Option A.

Note: A point on the screen where the wave crest of one wave falls on the wave crest of other, the resultant amplitude is maxima. A point on the screen where the wave crest of one falls on the wave trough of another, the resulting amplitude is minima.