Question

In young’s double slit experiment, the slits are separated by 0.56 mm and the screen is placed 2.8 cm away. The distance between the central bright fringe and the fifth bright fringe is 1.5 cm. Find the wavelength of light used.

Answer 1

When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark fringes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits. Regions of constructive interference, corresponding to bright fringes, are produced when the path difference from the two slits to the fringe is an integral number of wavelengths of the light. Destructive interference and dark fringes are produced when the path difference is a half-integral number of wavelengths.

Complete step by step solution:
$Y = \dfrac{{n\lambda D}}{d}$ where Y is the distance between the fringes.
And $\beta = \dfrac{{\lambda D}}{d}$is the distance between consecutive fringes.
Therefore, according to the question
$\begin{gathered} d = 0.56mm \\\ D = 2.8cm \\\ \beta 5 = 1.5cm \\\ \end{gathered}$
Here $\beta 5$ refers to the distance between the central bright fringe and the fifth bright fringe

\therefore this \Rightarrow \beta 5 = 5 \times \beta \\\ \Rightarrow \beta 5 = 5\dfrac{{\lambda D}}{d} \\\ \Rightarrow \lambda = \dfrac{{\beta 5 \times d}}{{5 \times D}} \\\ \Rightarrow \lambda = \dfrac{{1.5 \times 0.056}}{{5 \times 2.8}} \\\ \Rightarrow \lambda = 0.006cm \\\ \end{gathered} $$ **Hence the required wavelength of the light used will be $$\lambda = 0.006cm$$** **Note:** In solving cases of YDSE ( young’s double slit experiment) questions always be careful when dealing with fringes and always thoroughly the units to avoid any blunder.