Question

In an interference experiment, a third bright fringe is obtained at a point on the screen with a light of $700\;nm$. What should be the wavelength of the light source in order to obtain $5^{th}$ bright fringe at the same point?

Answer 1

Hint: We know that interference is the phenomena of superposition of waves which either result in the increase in the amplitude of the resultant wave or in the decrease in the amplitude of the resultant wave, hence we can solve the question as discussed below.

Formula used:
$b_n=\dfrac{n\lambda D}{d}$, where $b_n$ is the distance between the fringe and the centre, and $n$ is the number of bright fringes due to wavelength $\lambda$. Also $D$ is the distance between the screen and the slit, and $d$ is the distance between the slits.

Complete step-by-step solution:
Given that with wavelength $\lambda=700\;nm$, third bright fringe occurs at some point, then we have
Then $b_3=\dfrac{3\times 700 D}{d}=\dfrac{2100D}{d}$
Let the fifth fringe be formed at $b_5$ due to some wavelength $\lambda$ then we have
$b_5=\dfrac{5\lambda D}{d}$
Given tha $b_3=b_5$
Then, we have $\dfrac{2100D}{d}=\dfrac{5\lambda D}{d}$
$\implies 2100=5\lambda$
$\therefore \lambda=\dfrac{2100}{5}=420nm$
Hence the required answer is $420\;nm$
Additional information:
Interference is observed only when the source is coherent in nature, which means that the waves are at a constant phase difference and have only frequency. Also, the source must be almost point sources, and must be monochromatic in nature. The distance between the source and screen must be far to obtain visible fringes.
A very common example of fringes due to inference is the reflection of light due to soap bubbles, which results in interference.

Note: When the waves interfere constructively, we have the amplitude of the resultant wave to be greater than their parent individual waves and similarly, when the waves interfere destructively we have the amplitude of the resultant wave to be lesser than their parent individual waves.