In a young's double slit experiment, slits are illuminated by a monochromatic source of wavelength 6000 Å and fringes are obtained. If screen is moved by a distance of 5 cm towards slits, change in fringe width is $3 × 10^{–5} m$. Then separation between the slits will be
Answer & explanation
Correct answer: option 1
$β=\frac{λD}{d}$ and $β'=\frac{λD'}{d}$
$∴β-β'=\frac{λ(D-D')}{d}$
or $d=\frac{λ(D-D')}{(β-β')}$
$=\frac{6000×10^{-10}×5×10^{-2}}{3×10^{-5}}m=10^{-3}m=1mm$