Question

A beam of light consisting of two wavelengths, $650\,nm$ and $520 \,nm$ is used to obtain interference fringes in a Young?s double-slit experiment. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

• A. $1.17\, mm$
• B. $2.52\, mm$
• C. $1.56\, mm$
• D. $3.14\, mm$

Answer 1

Answer: (C)

$1.56\, mm$

Answer 2

Let at linear distance $'y'$ from center of screen the bright fringes due to both wavelength coincides. Let $n_1$ number of bright fringe with wavelength $\lambda_1$ coincides with $n_2$ number of bright fringe with wavelength $\lambda_2$. We can write $y = n_1 \beta_1 = n_2 \beta_2$ $n_1 \frac{\lambda_1 D}{d} = n_2 \frac{D\lambda_2}{d}$ or $n_1\lambda_1 = n_2\lambda_2\quad ...(i)$ Also at first position of coincide, the $n^{th}$ bright fringe of one will coincide with $(n + 1)^{th}$ bright fringe of other. lf $\lambda_2 < \lambda_1$, So, then $n_2 > n_1$ and $n_2 = n_1 + 1 \quad...(ii)$ Using equation $(ii)$ in equation $(i)$ $n_1\lambda_1 = (n_1 + 1) \lambda_2$ $n_1 (650) \times 10^{-9} = (n_1 + 1) 520 \times 10^{-9}$ $65 \,n_1 - 52 \,n_1 + 52$ or $13 \,n_1= 52$ or $n_1 = 4$ Thus, $y = n_1\beta_1$ $= 4[\frac{(6.5 \times 10^{-7})(1.2)}{2\times 10^{-3}}]$ $= 1.56 \times 10^{-3}\,m$ $= 1.56\,mm$ So, the fourth bright fringe of wavelength $520\, nm$ coincides with $5^{th}$ bright fringe of wavelength $650 \,nm$