- A$0.5$
- B$1.0$
- C$0.6$
- ✓$0.3$
$\lambda_{1}=5000 \,\mathring A$
$\beta_{1}=\frac{ D }{ d }\left(5000 \times 10^{-10}\right)=5 \times 10^{-4} \,m$......$(I)$
$\beta_{2}=\frac{ D }{(2 d )}\left(6000 \times 10^{-10}\right)= x \text { (let) }$.....$(II)$
Divide $(II)$ and $(I)$
$\frac{\beta_{2}}{\beta_{1}}=\frac{3000 \times 10^{-10}}{5000 \times 10^{-10}}=\frac{x}{5 \times 10^{-4}}$
$x=3 \times 10^{-4} m \text { or } 0.3 \,mm$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$y = a\sin (kx + \omega t)$ ......$(1)$
$y = a\sin (\omega t - kx)$ ......$(2)$
$y = a\cos (kx + \omega t)$ ......$(3)$
$y = a\cos (\omega t - kx)$ ......$(4)$
emitted by four different sources ${S_1},\,{S_2},\,{S_3}$ and ${S_4}$ respectively, interference phenomena would be observed in space under appropriate conditions when


Reason : Width of fringe is inversely proportional to the wavelength of the light used.

