Question

A beam of light consisting of wavelengths \(650\,\text{nm}\) and \(550\,\text{nm}\) illuminates Young’s double slits with separation \(d=2\,\text{mm}\) such that the interference fringes are formed on a screen placed at a distance \(D=1.2\,\text{m}\) from the slits. The least distance from the central maximum, where the bright fringes due to both wavelengths coincide, is ________ \(\times10^{-5}\,\text{m}\). Given: \[ \lambda_1=650\,\text{nm},\quad \lambda_2=550\,\text{nm},\quad d=2\times10^{-3}\,\text{m},\quad D=1.2\,\text{m} \]

Hint:

For coincidence of bright fringes in YDSE, always equate \(n_1\lambda_1=n_2\lambda_2\) and choose the smallest integers.

Answer 1

Correct Answer: 429

Concept: In Young’s double slit experiment, the position of the \(n\)-th bright fringe is: \[ y_n=\frac{n\lambda D}{d} \] Bright fringes for two wavelengths coincide when: \[ n_1\lambda_1=n_2\lambda_2 \]
Step 1: Find the least integers \[ 650n_1=550n_2 \Rightarrow 13n_1=11n_2 \] Smallest integers: \[ n_1=11,\quad n_2=13 \]
Step 2: Calculate the distance from central maximum \[ y=\frac{n_1\lambda_1 D}{d} =\frac{11\times650\times10^{-9}\times1.2}{2\times10^{-3}} \] \[ y=4.29\times10^{-3}\,\text{m} =429\times10^{-5}\,\text{m} \] Final Answer: \[ \boxed{429} \]