A beam of light consisting of two wavelengths 7000 $\overset{o}{A}$ and 5500 $\overset{o}{A}$ is used to obtain interference pattern in Young’s double slit experiment .The distance between the slits 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 x 10^-5m. The value of n is ______

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 462.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} n_{1} λ_{1} = n_{2} λ_{2} \\ n_{1} \times 7000 = n_{1} \times 5500 \\ n_{1} \times 70 = n_{2} \times 55 \\ n_{1} \times 14 = n_{2} \times 11 \end{array} \frac{n_{1}}{n_{2}} = \frac{11}{14}$
Position of central maxima
$\begin{array}{l} y = \frac{nλ D}{d} \\ y = \frac{11 \times 7000 \times 10^{- 10} \times 150 \times 10^{- 2}}{2.5 \times 10^{- 3}} \\ y = \frac{11 \times 7 \times 15 \times 10^{8} \times 10^{- 12}}{25} \\ y = 2.2 \times 21 \times 10^{- 4} \\ y = 22 \times 21 \times 10^{- 5} \\ y = 462 \times 10^{- 5} \\ n \times 10^{- 5} = 462 \times 10^{- 5} \\ \Rightarrow n = 462 \end{array}$