Question

Using Young's double slit experiment, a monochromatic light of wavelength $5000 \,\mathring{A}$ produces fringes of fringe width $0.5 \,mm$. If another monochromatic light of wavelength $6000 \,\mathring{A}$ is used and the separation between the slits is doubled, then the new fringe width will be :

• A. 0.5 mm
• B. 1.0 mm
• C. 0.6 mm
• D. 0.3 mm

Answer 1

Answer: (D)

Fringe width $\beta=\frac{ D \lambda}{ d }$
$\lambda_{1}=5000 \,\mathring{A} $
$\beta_{1}=\frac{ D }{ d }\left(5000 \times 10^{-10}\right)=5 \times 10^{-4} m \ldots $ (I)
$\beta_{2}=\frac{ D }{(2 d )}\left(6000 \times 10^{-10}\right)= x $ (let) ...(II)
Divide (II) & (I)
$\frac{\beta_{2}}{\beta_{1}}=\frac{3000 \times 10^{-10}}{5000 \times 10^{-10}}=\frac{ x }{5 \times 10^{-4}} $
$ x =3 \times 10^{-4} m $ or $ 0.3\, mm$