Question

A beam of light consisting of two wavelengths 7000 Å and 5500 Å is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is n × 10^-5 m. The value of n is_____.

Answer 1

Correct Answer: 462

Let \( n_1 \) maxima of 7000 \(\AA\) coincide with \( n_2 \) maxima of 5500 \(\AA\). Therefore, \( n_1 \beta_1 = n_2 \beta_2 \), \[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{5500}{7000} = \frac{11}{14} \] Hence, the 11th maximum of 7000 \(\AA\) coincides with the 14th maximum of 5500 \(\AA\). To find the least distance from this, \[ y = n_1 \beta_1 D / d \] \[ y = \frac{11 \times 7000 \times 10^{-10} \times 150 \times 10^{-2}}{2.5 \times 10^{-3}} = 462 \times 10^{-5} \, \text{m} \] Thus, \( n = 462 \)