In Young’s double slit experiment, two wavelengths λ1 = 780 nm and λ2 = 520 nm are used to obtain interference fringes. If the nth bright band due to λ1 coincides with (n + 1)th bright band due to λ2, then the value of n is

Question

In Young’s double slit experiment, two wavelengths λ1 = 780 nm and λ2 = 520 nm are used to obtain interference fringes. If the nth bright band due to λ1 coincides with (n + 1)th bright band due to λ2, then the value of n is (A) 4 (B) 3 (C) 2 (D) 6

Tardigrade Admin · Accepted Answer

xn = xn+1 n β1 = (n+1)β2 (nλ1D/d) =(n+1) (λ2D/d) n λ1 = (n+1)λ2 (n+1/n) =(λ1/λ2) = (780/520) 52n + 52 = 78 n 26 n = 52 n = (52/26) = 2

Q. In Young’s double slit experiment, two wavelengths $λ_{1} = 780 nm$ and $λ_{2} = 520 nm$ are used to obtain interference fringes. If the $n^{t h}$ bright band due to $λ_{1}$ coincides with $(n + 1)^{t h}$ bright band due to $λ_{2}$ , then the value of n is

4

3

2

6

$x_{n} = x_{n + 1}$
$^{n} β_{1} =^{(n + 1)} β_{2}$
$\frac{n λ _{1} D}{d} = (n + 1) \frac{λ _{2} D}{d}$
$n λ_{1} = (n + 1) λ_{2}$
$\frac{n + 1}{n} = \frac{λ _{1}}{λ _{2}} = \frac{780}{520}$
$52 n + 52 = 78 n$
$26 n = 52$
$n = \frac{52}{26} = 2$

Q. In Young’s double slit experiment, two wavelengths λ1​=780nm and λ2​=520nm are used to obtain interference fringes. If the nth bright band due to λ1​ coincides with (n+1)th bright band due to λ2​, then the value of n is

4

3

2

6

xn​=xn+1​ nβ1​=(n+1)β2​ dnλ1​D​=(n+1)dλ2​D​ nλ1​=(n+1)λ2​ nn+1​=λ2​λ1​​=520780​ 52n+52=78n 26n=52 n=2652​=2

Q. In Young’s double slit experiment, two wavelengths $λ_{1} = 780 nm$ and $λ_{2} = 520 nm$ are used to obtain interference fringes. If the $n^{t h}$ bright band due to $λ_{1}$ coincides with $(n + 1)^{t h}$ bright band due to $λ_{2}$ , then the value of n is

$x_{n} = x_{n + 1}$
$^{n} β_{1} =^{(n + 1)} β_{2}$
$\frac{n λ _{1} D}{d} = (n + 1) \frac{λ _{2} D}{d}$
$n λ_{1} = (n + 1) λ_{2}$
$\frac{n + 1}{n} = \frac{λ _{1}}{λ _{2}} = \frac{780}{520}$
$52 n + 52 = 78 n$
$26 n = 52$
$n = \frac{52}{26} = 2$