Question 13 Marks
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
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide?
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide?
Answer
View full question & answer→Here, $\lambda_1=650 nm=650 \times 10^{-9} m$,
$\lambda_2=520 nm=520 \times 10^{-9} m$
Suppose $\quad d=$ distance between two slits,
$D =$ distance of screen from the slits.
(a) For third bright fringe, n=3
$\therefore \quad x=n \lambda_1 \frac{ D }{d}=3 \times 650 \frac{ D }{d} nm=1950 \frac{ D }{d} mm$
(b) Let $n$th fringe due to $\lambda_2=520 nm$ coincide with $(n-1)$ th bright fringe due to $\lambda_1=650 nm$.
$n \lambda_2=(n-1) \lambda_1$
$n \times 520=(n-1) 650$
$4 n=5 n-5$ or $n=5$.
$\therefore$ The least distance required, $x=n \lambda_2 \frac{ D }{d}$
$=5 \times 520 \frac{ D }{d}=2600 \frac{ D }{d} nm$.
$\lambda_2=520 nm=520 \times 10^{-9} m$
Suppose $\quad d=$ distance between two slits,
$D =$ distance of screen from the slits.
(a) For third bright fringe, n=3
$\therefore \quad x=n \lambda_1 \frac{ D }{d}=3 \times 650 \frac{ D }{d} nm=1950 \frac{ D }{d} mm$
(b) Let $n$th fringe due to $\lambda_2=520 nm$ coincide with $(n-1)$ th bright fringe due to $\lambda_1=650 nm$.
$n \lambda_2=(n-1) \lambda_1$
$n \times 520=(n-1) 650$
$4 n=5 n-5$ or $n=5$.
$\therefore$ The least distance required, $x=n \lambda_2 \frac{ D }{d}$
$=5 \times 520 \frac{ D }{d}=2600 \frac{ D }{d} nm$.
