In Young's experiment, the third bright band for light of wavelength 6000 mathringA coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is

Question

In Young's experiment, the third bright band for light of wavelength 6000 mathringA coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is (A)  3600 mathringA  (B)  4000 mathringA  (C)  5000 mathringA  (D)  4500 mathringA

Tardigrade Admin · Accepted Answer

x=(mDλ 1/d)=((m+1)Dλ 2/d) ⇒ 3× 6000=4λ 2 Or λ 2=(3× 6000/4) =4500 mathringA

Q. In Young's experiment, the third bright band for light of wavelength $6000 \overset{˚}{A}$ coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is

$3600 \overset{˚}{A}$

$4000 \overset{˚}{A}$

$5000 \overset{˚}{A}$

$4500 \overset{˚}{A}$

$5500 \overset{˚}{A}$

$x = \frac{m D λ _{1}}{d} = \frac{( m + 1 ) D λ _{2}}{d}$
$\Rightarrow$ $3 \times 6000 = 4 λ_{2}$
Or $λ_{2} = \frac{3 \times 6000}{4}$
$= 4500 \overset{˚}{A}$

Q. In Young's experiment, the third bright band for light of wavelength 6000A˚ coincides with the fourth bright band for another source of light in the same arrangement. Then the wavelength of second source is

3600A˚

4000A˚

5000A˚

4500A˚

5500A˚

x=dmDλ1​​=d(m+1)Dλ2​​ ⇒ 3×6000=4λ2​ Or λ2​=43×6000​ =4500A˚